Calculus 1 Exam 2 Review

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Problem 1

Define the secant and tagent slopes to a curve.  Define average and instantaneous velocity.

Solution

A secant slope to a curve is the slope of the line joining 2 distinct points on the curve.  In terms of a formula slope of the secant line for y=f(x) through points corresponding to x=a and x=b is

({{({{(Subscript[m, sec])=((f(b)-f(a))/(b-a))=((f(a+h)-f(a))/h), where h=b-a}})}})

The tangent slope to a curve is the slope of the tangent line at a particular point x=a.  The tangent line is the line through the given point which goes in the “same direction” as the curve.  We define tangent slopes as the limit of secant slopes, yielding two equivalent-but-different-looking formulas:

({{({{(Subscript[m, tan])=(Underscript[lim , b->a])}})((f(b)-f(a))/(b-a))=(Underscript[lim , h->0])((f(a+h)-f(a))/h)}})

The first form is often used for estimation purposes.  Many students prefer the second form as the h->0 part is always the same no matter at what point on the curve you are computing tangent slopes for, and f(a+h) often requires multiplying something out, which many people find algebraically easier than the factoring that often occurs in the first version.

Problem 2

We have not (and won’t until Calculus 2) define the function ln(x) which appears on most calculators.  Estimate the slope of the tangent line to y=ln(x) at x=2.

Solution

If we use the first version of the tangent slope formula we would be estimating (lim_( b->2))((ln(b)-ln(2))/(b-2)).  Using a standard limit-estimation table from Chapter 1 we then have

NumericalLimitByDelta[( Log[b]-Log[2])/(b-2),{b,2}]

(Note:  Mathematica uses the symbol “log” for what on a calculator appears as “ln”, which can be a bit confusing as there is also a calculator symbol “log” which means something different)

Based on this table it looks like the tangent slope to ln(x) at x=2 is 0.5, or (1/2).

Problem 3

Using limits (and without deriving a general formula), find the tangent lines to y=2(x^2)-3x+5 at x=1 and x=2.

Solution

For x=1 we have that y=2 (1^2)-3 1+5=4, so the curve point is (1,4).  The tangent slope is given by

({{({{(Subscript[m, tan])=(Underscript[lim , h->0])(((2(((1+h))^2)-3(1+h)+5)-4)/h)=(Underscript[lim , h->0])(((2(1+2h+(h^2))-3-3h+5)-4)/h)=}, {(Underscript[lim , h->0])((2+4h+2(h^2)-3-3h+5-4)/h)=(Underscript[lim , h->0])((h+2(h^2))/h)=(Underscript[lim , h->0])1+2h=1}})}})

So the tangent line at x=1 is the line through (1,4) with slope 1.  Using the point slope-form for a line ( y-(Subscript[y, 0])=m(x-(Subscript[x, 0])) ) the line is y-4=1(x-1), or y=x+3.

For x=2 we have that y=2 (2^2)-3 2+5=7, so the curve point is (2,7).  Then tangent slope at this point is given by

({{({{(Subscript[m, tan])=(Underscript[lim , h->0])(((2(((2+h))^2)-3(2+h)+5)-4)/h)=(Underscript[lim , h->0])(((2(4+4h+(h^2))-6-3h+5)-7)/h)=}, {(Underscript[lim , h->0])((8+8h+2(h^2)-6-3h+5-7)/h)=(Underscript[lim , h->0])((5h+2(h^2))/h)=(Underscript[lim , h->0])5+2h=5}})}})

So the tangent line at x=2 is the line through (2,7) with slope 5.  Using the point-slope for a line we get y-7=5(x-2), or y=5x-3.

Problem 4

Suppose that f(x) is a function and rather than an actual formula for the function you only have a list of isolated points on the graph of y=f(x) - say (1,3), (1.8, 3.9), (2,4), (2.5, 4.1), and (3,5).  How could use estimate f'(2)?

Solution

There is no precise way to answer this problem, but we can come up with a fairly decent estimate using the idea that the tangent slope at a point is the limit of secant slopes.  By looking at several secant slopes based on points “near” x=2 we might be able to average them to get better estimates:

Slope from (1.8,3.9) to (2,4): m=((4-3.9)/(2-1.8))=(0.1/0.2)=.1
Slope from (2,4) to  (2.5,4.1): m=((4.1-4)/(2.5-2))=((.1)/(.5))=.2
Slope from (1.8,3.9) to (2.5,4.1): m=((4.1-3.9)/(2.5-1.8))=((.2)/(.7))=(2/7)≈0.286

Averaging these three values gives an estimate of ((.1+.2+.286)/3)≈0.195, so we can use that as an estimate for the tangent slope at x=2.  You’ll note that I never used the first or last point - they are both 1 unit away from x=2 which might be too far away to yield useful information for the tangent line at x=2.

Problem 5

State both definitions of the derivative f'(x) (the “2 point” and “1 point” definitions).

Solution

({{({{f'(x)=(Underscript[lim , b->x])((f(x)-f(b))/(x-b))}, {f'(x)=(Underscript[lim , h->0])((f(x+h)-f(x))/h)}})}})

Both forms always yield the same result; many (but not all!) students find the second form easier to use as for algebriac functions f(x) the factoring is typically easier.  note that in the first definition the numerator and denominator can both be reversed to give the same result.

Problem 6

Use both definitions of the derviative to find f'(x) for f(x)=(x^3)-3x+2.

Solution

Using the “two point” definition:

({{({{f'(x)=(Underscript[lim , b->x])((f(x)-f(b))/(x-b))=}})(Underscript[lim , b->x])((((x^3)-3x+2)-((b^3)-3b+2))/(x-b))=(Underscript[lim , b->x])(((x^3)-3x+(3b-(b^3)))/(x-b))}})

Using long division you can verify that (x^3)-3x+(3b-(b^3))=(x-b)((x^2)+b x+(b^2)-3), so

({{({{(Underscript[lim , b->x])(((x^3)-3x+(3b-(b^3)))/(x-b))=(Underscript[lim , b->x])(((x-b)((x^2)+b x+(b^2)-3))/(x-b))=(Underscript[lim , b->x])}})(x^2)+b x+(b^2)-3}, {=(x^2)+x(x)+(((x))^2)-3=3(x^2)-3}})

So if f(x)=(x^3)-3x+2 then f'(x)=3(x^2)-3.

Using the “one point” definition:

({{({{f'(x)=(Underscript[lim , h->0])}})((f(x+h)-f(x))/h)=(Underscript[lim , h->0])((((((x+h))^3)-3(x+h)+2)-((x^3)-3x+2))/h)=}, {(Underscript[lim , h->0])((((x^3)+3(x^2)h+3x (h^2)+(h^3)-3x-3h+2)-((x^3)-3x+2))/h)=}, {(Underscript[lim , h->0])((3(x^2)h+3x (h^2)+(h^3)-3h)/h)=(Underscript[lim , h->0])((h(3(x^2)+3x h+(h^2)-3))/h)=}, {(Underscript[lim , h->0])3(x^2)+3x h+(h^2)-3=3(x^2)+3x(0)+(0^2)-3=3(x^2)-3}})

So if f(x)=(x^3)-3x+2 then f'(x)=3(x^2)-3.

Note: The Pascal’s triangle-assisted expansion of (((x+h))^3) in the second approach is easier than the long division in the first method, which is why in these problems many people use the second version.

Problem 7

Use a definition of the derivative to find f'(x) for f(x)=(1/(2x-3)).

Solution

({{({{({{f'(x)=(Underscript[lim , h->0])}})((f(x+h)-f(x))/h)=(Underscript[lim , h->0])}})(((1/(2(x+h)-3))-(1/(2x-3)))/h)=(Underscript[lim , h->0])(1/h)((1/(2x+2h-3))-(1/(2x-3)))=}, {(Underscript[lim , h->0])(1/h)(((2x-3)/((2x+2h-3)(2x-3)))-((2x+2h-3)/((2x+2h-3)(2x-3))))=(Underscript[lim , h->0])(1/h)(((-2h)/((2x+2h-3)(2x-3))))=}, {(Underscript[lim , h->0])((-2)/((2x+2h-3)(2x-3)))=((-2)/((2x+2 0-3)(2x-3)))=-(2/(((2x-3))^2))}})

So if f(x)=(1/(2x-3)), f'(x)=-(2/(((2x-3))^2))

Problem 8

Use a definition of the derivative to find f'(x) for f(x)=(Sqrt[9x+1])

Solution

({{({{({{({{f'(x)=(Underscript[lim , h->0])}})((f(x+h)-f(x))/h)=(Underscript[lim , h->0])}})(((Sqrt[9(x+h)+1])-(Sqrt[9x-1]))/h)}})=(Underscript[lim , h->0])(((Sqrt[9x+9h+1])-(Sqrt[9x+1]))/h)=}, {(Underscript[lim , h->0])(((Sqrt[9x+9h+1])-(Sqrt[9x+1]))/h)×(((Sqrt[9x+9h+1])+(Sqrt[9x+1]))/((Sqrt[9x+9h+1])+(Sqrt[9x+1])))=(Underscript[lim , h->0])((((((Sqrt[9x+9h+1])))^2)-((((Sqrt[9x+1])))^2))/(h((Sqrt[9x+9h+1])+(Sqrt[9x+1]))))=}, {(Underscript[lim , h->0])(((9x+9h+1)-(9x+1))/(h((Sqrt[9x+9h+1])+(Sqrt[9x+1]))))=(Underscript[lim , h->0])((9h)/(h((Sqrt[9x+9h+1])+(Sqrt[9x+1]))))=}, {(Underscript[lim , h->0])(9/(((Sqrt[9x+9h+1])+(Sqrt[9x+1]))))=(9/((Sqrt[9x+9 0+1])+(Sqrt[9x+1])))=(9/(2(Sqrt[9x+1])))}})

So if f(x)=(Sqrt[9x+1]), f'(x)=(9/(2(Sqrt[9x+1]))).

Problem 9

Use a definition of the derivative to find f'(x) for f(x)=2(x^4)-6(x^2)+5145

Solution

({{({{({{({{({{f'(x)=(Underscript[lim , h->0])}})((f(x+h)-f(x))/h)=(Underscript[lim , h->0])}})(((2(((x+h))^4)-6(((x+h))^2)+5145)-(2(x^4)-6(x^2)+5145))/h)=}, {(Underscript[lim , h->0])(((2((x^4)+4(x^3)h+6(x^2)(h^2)+4x (h^3)+(h^4))-6((x^2)+2x h+(h^2))+5145)-(2(x^4)-6(x^2)+5145))/h)=}, {(Underscript[lim , h->0])((2(x^4)+8(x^3)h+12(x^2)(h^2)+8 x (h^3)+2(h^4)-6(x^2)-12x h-6(h^2)+5145-2(x^4)+6(x^2)+5145)/h)=}, {(Underscript[lim , h->0])((8(x^3)h+12(x^2)(h^2)+8 x (h^3)+2(h^4)-12x h-6(h^2))/h)=(Underscript[lim , h->0])((h(8(x^3)+12(x^2)h+8 x (h^2)+2(h^3)-12x -6h))/h)=}, {(lim_(h->0))8(x^3)+12(x^2)h+8 x h+2(h^3)-12x-6h=}, {8(x^3)+12(x^2)×0+8x×0+2×(0^3)-12x-6 0 =}, {8(x^3)-12x}})}})}})

Where (((x+h))^4)=(x^4)+4(x^3)h+6(x^2)(h^2)+4x (h^3)+(h^4) by Pascal’s triangle.

So if f(x)=2(x^4)-6(x^2)+5145, then f'(x)=8(x^3)-12x.

Problem 10

Use a definition of the derivative to find f'(x) for f(x)=(x^(1/3)). (Note: this is algebraically more complicated, as it requires a generalization of the conjugate trick - think of the conjugate trick as a clever use of the identity (a+b)(a-b)=(a^2)-(b^2); for this problem try the related identity (a-b)((a^2)+a b+(b^2))=(a^3)-(b^3))

Solution

({{({{({{({{({{f'(x)=(Underscript[lim , h->0])}})((f(x+h)-f(x))/h)=(Underscript[lim , h->0])}})((((x+h)^(1/3))-(x^(1/3)))/h)}})=}, {(Underscript[lim , h->0])((((x+h)^(1/3))-(x^(1/3)))/h)×(((((((x+h)^(1/3))))^2)+((x+h)^(1/3))(x^(1/3))+((((x^(1/3))))^2))/((((((x+h)^(1/3))))^2)+((x+h)^(1/3))(x^(1/3))+((((x^(1/3))))^2)))=}, {(Underscript[lim , h->0])(((((((x+h)^(1/3))))^3)+(((((x+h)^(1/3))))^2)(x^(1/3))+((x+h)^(1/3))((((x^(1/3))))^2)-(((((x+h)^(1/3))))^2)(x^(1/3))-((x+h)^(1/3))((((x^(1/3))))^2)-((((x^(1/3))))^3))/(h((((((x+h)^(1/3))))^2)+((x+h)^(1/3))(x^(1/3))+((((x^(1/3))))^2))))=}, {(Underscript[lim , h->0])(((((((x+h)^(1/3))))^3)-((((x^(1/3))))^3))/(h((((((x+h)^(1/3))))^2)+((x+h)^(1/3))(x^(1/3))+((((x^(1/3))))^2))))=(Underscript[lim , h->0])(((x+h)-x)/(h((((((x+h)^(1/3))))^2)+((x+h)^(1/3))(x^(1/3))+((((x^(1/3))))^2))))=}, {(Underscript[lim , h->0])(h/(h((((((x+h)^(1/3))))^2)+((x+h)^(1/3))(x^(1/3))+((((x^(1/3))))^2))))=(Underscript[lim , h->0])(1/((((((x+h)^(1/3))))^2)+((x+h)^(1/3))(x^(1/3))+((((x^(1/3))))^2)))=}, {(1/((((((x+0)^(1/3))))^2)+((x+0)^(1/3))(x^(1/3))+((((x^(1/3))))^2)))=(1/(((((x^(1/3))))^2)+(x^(1/3))(x^(1/3))+((((x^(1/3))))^2)))=(1/(3((((x^(1/3))))^2)))}})}})

So if f(x)=(x^(1/3)), then f'(x)=(1/(3((((x^(1/3))))^2))) (which should make you appreciate the power rule - (d/(dx))(x^(1/3))=(1/3)(x^(-2/3))=(1/(3((((x^(1/3))))^2)))!!!)

Problem 11

Solution

If f(x)=(x^2)-3x+1, find the tangent line to the curve y=f(x) at x=1, using the definition of the derivative to find the slope.

({{({{({{({{f'(x)=(Underscript[lim , h->0])}})((f(x+h)-f(x))/h)=(Underscript[lim , h->0])}})((((((x+h))^2)-3(x+h)+1)-((x^2)-3x+1))/h)}})=}, {({{({{(Underscript[lim , h->0])}})((((x^2)+2x h+(h^2)-3x-3h+1)-((x^2)-3x+1))/h)}})=({{(Underscript[lim , h->0])}})((2x h+(h^2)-3h)/h)=}, {({{(Underscript[lim , h->0])}})((h(2x +h-3))/h)=({{(Underscript[lim , h->0])}}) 2x+h-3=2x-3}})

So the tangent slope is f'(1)=2 1-3=-1.  f(1)=(1^2)-3 1+1=-1, so the point in question is (1,-1).  Therefore in point-slope form the tangent line is y-(-1)=(-1)(x-1), or y=-x.

Problem 12

The temperature during a given day is given by f(t)=74+18 sin((π/12)t), where t is in hours after midnight and temperature is measured in Fahrenheit.  How fast is the temperature changing at 8 p.m.?  At 4 p.m.?

Solution

The rate of change of temperature is given by f'(t), and we have

({{({{f'(t)=(d/(dt))(74+18sin((π/12)t))=0+18cos((π/12)t)×(π/12)=((3π)/2)}})cos(((π t)/12))}})

The time 8 p.m. corresponds to t=20, and f'(20)=((3π)/2)cos(((20π)/12))=((3π)/2)cos(((5π)/3))=((3π)/2)×(1/2)=((3π)/4) degrees Fahrenheit per hour.

The time 4 p.m. corresponds to t=16, and f'(16)=((3π)/2)cos(((16π)/12))=((3π)/2)cos(((4π)/3))=((3π)/2)×-(1/2)=-((3π)/4) degrees Fahrenheit per hour.

Problem 13

A ball is tossed upwards off a cliff.  At any time t the height of the ball above the cliff is given by f(t)=4+256t-16(t^2) feet, where t is measured in seconds.  How fast is the ball moving after 3 seconds?  After 10 seconds?  When is the ball motionless?  What happens to the velocity of the ball in the long run, and is this realistic?  What is the acceleration of the ball at any point in time?

Solution

The velocity and acceleration of the ball are given by the first and second derivatives respectively, so v=f'(t)=(d/(dt))(4+256t-16(t^2))=256-32t feet per second and a=f''(t)=(d/(dt))v=(d/(dt))(256-32t)=-32 feet per second squared (so the acceleration does not depend on time).  So after 3 seconds the velocity of the ball is v(3)=256-32 3=160 feet per second; after 10 seconds the velocity is v(10)=256-32 10=-64 feet per second (in this case the ball is headed back down).  The question of when the ball is motionless is that of solving v(t)=0, and 256-32t=0 has the solution t=(256/32)=8 seconds.  The question of the long term velocity is the compution (Underscript[lim , t->∞])v(t)=(Underscript[lim , t->∞])(256-32t)=-∞.  This isn’t realistic for two reasons.  First as the ball is falling faster and faster it must reach the base of the cliff (no matter how tall the cliff is) at some point, and when it hits the base the velocity is going to change dramatically.  In addition even if the cliff was bottomless the velocity would increase to the point where air resistance would play in an important role in how the velocity changes over time, in which case the formula v(t)=256-32t is no longer a realistic one.

Problem 14

Find the tangent lines to the curve y=(x/(2x+3)) at x=0 and x=1.  What happens to the tangent slopes as x->-(3/2)?

Solution

At x=0 and x=1 we have the points (0,0) and (1,(1/5)) respectively.  The tangent slopes are given by y'=(d/(dx)) (x/(2x+3))=(((2x+3)(d/(dx))x-x(d/(dx))(2x+3))/(((2x+3))^2))=(((2x+3)×1-x×2)/(((2x+3))^2))=(3/(((2x+3))^2)).  So at x=0 the tangent slope is y'(0)=(3/(3^2))=(1/3) and at x=1 the tangent slope is y'(1)=(3/(5^2))=(3/25).  Therefore the tangent line at x=0 is y-0=(1/3)(x-0) (or  y=(1/3)x) and at x=1 it is y-(1/5)=(3/25)(x-1) (or y=(3/(25 ))x+(2/25)).

(Underscript[lim , x->-(3/2)])y'=(Underscript[lim , x->-(3/2)])(3/(((2x+3))^2))=∞ (in fact it is ∞ using limits from either side), so the tangent lines approach vertical as x->-(3/2).

Problem 15

Find the exact value of the limit (Underscript[lim , t->0])(((((1+t))^1000)-1)/t) by recognizing it as a derivative.

Solution

Pattern matching the given  limit against ({{f'(x)=(Underscript[lim , h->0])}})((f(x+h)-f(x))/h) it appears that “h” is being represented by t, the value x+h by 1+t (so x=1), and the basic function expression is f(x)=(x^1000) (note that f(1)=(1^1000)=1 which is necessary to match the pattern at the end of the numerator).  So the given limit is the derviative of (x^1000) at x=1.  The derivative of (x^1000) is 1000(x^999), and substituting in x=1 gives that the limit is 1000.

Problem 16

Draw a graph for a function y=f(x) which is not differentiable at 3 points, each time for a different reason.

Plot[ Piecewise[ {{1,x<1}, {-1, x<1}}]+ CubeRoot[x-3]+Abs[x-5], {x,-3,7}, Exclusions->x==1, Ticks-> {Range[-10,10], Range[-10,10]}]

Solution

Note this has discontinuity at x=1 (and not continuous => not differentiable), a vertical tangent at x=3 (so the derivative is not a number), and a sharp corner at x=5 (the quantity that defines the derivative is not the same as you come into 5 from the left as it is as when you come in from the right - so the two-sided limit that defined the derivative does not exist).

Problem 17

Draw a graph which is continuous but not differentiable.  Draw a graph which is differentiable but not continuous.

Solution

The graph of absolute value is continuous at x=0 but not differentiable there:

Plot[Abs[x], {x,-2,2}]

A differentiable but not continuous graph is impossible, as differentiability implies continuity.

Problem 18

What are all the derivative rules we have (including the “algebraic”-style rules)?

Solution

Sum rule: (d/(dx))(f(x)+g(x))=(d/(dx))f(x)+(d/(dx))g(x)
Difference rule: (d/(dx))(f(x)-g(x))=(d/(dx))f(x)-(d/(dx))g(x)
Constant multiple rule: If c is a real number, (d/(dx))c f(x)=c (d/(dx))f(x)
Product Rule: (d/(dx))( f(x)g(x) )=f(x)(d/(dx))g(x)+g(x)(d/(dx))f(x)
Quotient Rule: (d/(dx))((f(x))/(g(x)))=((g(x)(d/(dx))f(x)-f(x)(d/(dx))g(x))/((g(x))^2))
Chain Rule: (d/(dx))f(g(x))=f'(g(x))(d/(dx))g(x)

Power rule:  If r is a real number, then (d/(dx))(x^r)=r (x^(r-1)).
	Chain rule version: (d/(dx))(u^r)=r (u^(r-1)) ((du)/(dx))
Trigonometric Rules (in radian measure; λ represents a fixed real number; u any function):
(d/(dx))sin(x)=cos(x); (d/(dx))sin(λ x)=λ cos(λ x); (d/(dx))sin(u)=cos(u)((du)/(dx))
(d/(dx))cos(x)=-sin(x); (d/(dx))cos(λ x)=-λ sin(λ x); (d/(dx))cos(u)=-sin(u)((du)/(dx))
(d/(dx))tan(x)=(sec^2)(x); (d/(dx))tan(λ x)=λ (sec^2)(λ x); (d/(dx))tan(u)=(sec^2)(u)((du)/(dx))
(d/(dx))cot(x)=-(csc^2)(x); (d/(dx))cot(λ x)=-λ (csc^2)(λ x); (d/(dx))cot(u)=-(csc^2)(u)((du)/(dx))
(d/(dx))sec(x)=sec(x)tan(x);(d/(dx))sec(λ x)=λ sec(λ x)tan(λ x);(d/(dx))sec(u)=sec(u)tan(u)((du)/(dx))
(d/(dx))csc(x)=-csc(x)cot(x);(d/(dx))csc(λ x)=-λ csc(λ x)csc(λ x);(d/(dx))csc(u)=-csc(u)cot(u)((du)/(dx))

Problem 19

What are all of the different notations used for the derivative of y=f(x)?

Solution

({{({{f'(x)=y'=((dy)/(dx))=(d/(dx))f(x)=(d/(dx))y}})}})

There is also an older notation used by Isaac Newton based on “fluxions”, (OverDot[y]), but it is rarely used.

Problem 20

Find the derivative of (x^3)-3(x^2)+2x+1 with respect to x.

Solution

({{({{(d/(dx))}})((x^3)-3(x^2)+2x+1)=(d/(dx))(x^3)-3(d/(dx))(x^2)+2(d/(dx))x+(d/(dx))1=}, {3(x^2)-3(2x)+2(1(x^0))+0=3(x^2)-6x+2}})

Problem 21

Find the derivative of f(t)=3(t^5)-15t+5 with respect to t

Solution

({{({{(d/(dt))}})(3(t^5)-15t+5)=3(d/(dt))(t^5)-15 (d/(dt))t+(d/(dt))5 =}, {3(5(t^4))-15(1(t^0))+0=15(t^4)-15}})

Problem 22

Find the derivative of f(x)=2(x^2)-3x+(5/x)-2(Sqrt[x]) with respect to x

Solution

It is best to rewrite the expression using fractional and negative powers before we take the derivative (so we can use the power rule):

({{({{(d/(dx))(2(x^2)-3x+(5/x)-2(Sqrt[x]))=(d/(dx))(2(x^2)-3x+5(x^(-1))-2(x^(1/2)))}})=}, {2(d/(dx))(x^2)-3(d/(dx))x+5(d/(dx))(x^(-1))-2(d/(dx))(x^(1/2))=2(2x)-3 1+5(-1(x^(-2)))-2((1/2)(x^(-1/2)))=}, {4x-3-(5/(x^2))-(1/(Sqrt[x]))}})

Problem 23

Find (d/(da))(a (x^2)+b x+c)

Solution

This is kind of trick question - note the variable that the derivative is with respect to (a, which means all other letters shouldbe treated as constants):

({{({{(d/(da))(a (x^2)+b x+c)}})=(x^2)(d/(da))a+b x(d/(da))1+(d/(da))c=(x^2)×1+b x×0+0=(x^2)}})

Had the derivative been with respect to x the result would have been 2 a x+b.

Problem 24

Find (d/(dx))(tan(x)-(x/(sin(x))))

Solution

({{({{(d/(dx))(tan(x)-(x/(sin(x))))=(d/(dx))tan(x)-(d/(dx))(x/(sin(x))))=(sec^2)(x)-(d/(dx)) (x/(sin(x)))}})=}, {(sec^2)(x)-((sin(x)(d/(dx))x-x (d/(dx))sin(x))/((sin^2)(x)))=(sec^2)(x)-((sin(x)×1-x×cos(x))/((sin^2)(x)))=}, {(sec^2)(x)-((sin(x)-x cos(x))/((sin^2)(x)))}})

Problem 25

Find (d/(dx))(7^8)

Solution

Trick question - (7^8) is a constant, so (d/(dx))(7^8)=0.  This is meant to trick you into using the power rule to get the wrong answer 8×(7^7) (the power rule combined with the chain rule does work:  (d/(dx))(7^8)=8×(7^7)(d/(dx))7=8×(7^7)×0=0)

Problem 26

Find the derivative of (((x^2)-3x+1)/(2x+7)) with respect to x

Solution

({{({{(d/(dx))}})((((x^2)-3x+1)/(2x+7)))=(((2x+7) ({{(d/(dx))}})((x^2)-3x+1)-((x^2)-3x+1)({{(d/(dx))}})(2x+7))/(((2x+7))^2))=}, {(((2x+7)(2x-3)-((x^2)-3x+1)×2)/(((2x+7))^2))=(((4(x^2)+8x-21)-(2(x^2)-6x+2))/(((2x+7))^2))=}, {((2(x^2)+14x-23)/(((2x+7))^2))}})

Problem 27

Find ((dy)/(dx)) if y=sin(x)cot(x)+tan(x)

Solution

({{({{y'=(d/(dx))(sin(x)cot(x)+tan(x))=sin(x)}})(d/(dx))cot(x)+cot(x)(d/(dx))sin(x)+(d/(dx))tan(x)=}, {sin(x)(-(csc^2)(x))+cot(x)(cos(x))+(sec^2)(x)=}, {-csc(x)+cos(x)cot(x)+(sec^2)(x)}})

This is not the best way to do this problem however.  As cot(x)=((cos(x))/(sin(x))) y has the simpler equation y=cos(x)+tan(x), so y'=-sin(x)+(sec^2)(x) (some trignometric identities will convert one form to the other).

Problem 28

Find the derivative of ((1+cot(x))/(1-tan(x))) with respect to x

Solution

({{({{(d/(dx))}})((1+cot(x))/(1-tan(x)))=(((1-tan(x))({{(d/(dx))}})(1+cot(x))-(1+cot(x))({{(d/(dx))}})(1-tan(x)))/(((1-tan(x)))^2))=}, {(((1-tan(x))(-(csc^2)(x))-(1+cot(x))(-(sec^2)(x)))/(((1-tan(x)))^2))=((-(csc^2)(x)+(csc^2)(x)tan(x)+(sec^2)(x)+cot(x)(sec^2)(x))/(((1-tan(x)))^2))=}, {((-(csc^2)(x)+csc(x)sec(x)+(sec^2)(x)+csc(x)sec(x))/(((1-tan(x)))^2))=((-(csc^2)(x)+2sec(x)csc(x)+(sec^2)(x))/(((1-tan(x)))^2))}})

Problem 29

Find the derivative of sec(3x)-(x^5)csc(5x) with respect to x

Solution

({{({{(d/(dx))}})sec(3x)-(x^5)csc(5x)=({{(d/(dx))}})sec(3x)-({{(d/(dx))}})(x^5)csc(5x)=}, {sec(3x)tan(3x)({{(d/(dx))}})3x-((x^5)({{(d/(dx))}})csc(5x)+csc(5x)({{(d/(dx))}})(x^5))=}, {sec(3x)tan(3x)×3-((x^5)(-csc(5x)cot(5x)({{(d/(dx))}})5x)+csc(5x)(5(x^4)))=}, {3sec(3x)tan(3x)+(x^5)csc(5x)cot(5x)×5-5(x^4)csc(5x)=}, {3sec(3x)tan(3x)+5(x^5)csc(5x)cot(5x)-5(x^4)csc(5x)}})

Problem 30

Find f'(x) if f(x)=(x^3)cos(3x)sin(2x).

Solution

To approach this triple product think of the cosine and sine together as a single function:

({{({{(d/(dx))(x^3)cos(3x)sin(2x)=}})(d/(dx))(x^3)(cos(3x)sin(2x))=}, {(x^3)(d/(dx))(cos(3x)sin(2x))+(cos(3x)sin(2x))(d/(dx))(x^3)=}, {(x^3)(cos(3x)(d/(dx))sin(2x)+sin(2x)(d/(dx))cos(3x))+3(x^2)cos(3x)sin(2x)=}, {(x^3)(cos(3x)cos(2x)×2+sin(2x)(-sin(3x)×3))+3(x^2)cos(3x)sin(2x)=}, {2(x^3)cos(3x)cos(2x)-3(x^3)sin(2x)sin(3x)+3(x^2)cos(3x)sin(2x)}})

Note:  You can use this idea to develop your own “triple product rule” - (d/(dx))f(x)g(x)h(x)=((d/(dx))f(x))g(x)h(x)+f(x)((d/(dx))g(x))h(x)+f(x)g(x)(d/(dx))h(x).

Problem 31

Find the derivative of (Sqrt[1+(sin^2)(x)]) with respect to x

Solution

Rewriting this as f(x)=(((1+(sin^2)(x)))^(1/2)) to take advantage of the generalized power rule gives

({{({{(d/(dx))}})(((1+(sin^2)(x)))^(1/2))=(1/2)(((1+(sin^2)(x)))^(-1/2))({{(d/(dx))}})(1+(sin^2)(x))=}, {(1/2)(((1+(sin^2)(x)))^(-1/2))(0+2 (sin^1)(x)(d/(dx))sin(x))=}, {(1/2)(((1+(sin^2)(x)))^(-1/2))(2 sin(x)cos(x))=((sin(x)cos(x))/(Sqrt[1+(sin^2)(x)]))}})

Problem 32

Find the derivative of (((1+(1/x)))^23) with respect to x

Solution

Rewriting this as (((1+(1/x)))^23)=(((1+(x^(-1))))^23)  gives

({{({{({{(d/(dx))}})(((1+(x^(-1))))^23)=23(((1+(x^(-1))))^22)({{(d/(dx))}})(1+(x^(-1)))=23(((1+(x^(-1))))^22)(0+(-1)(x^(-2)))=}, {23(((1+(1/x)))^22)×((-1)/(x^2))=-(23/(x^2))(((1+(1/x)))^22)}})}})

Problem 33

Find the derivative of ((((1+sin(3t)))^4)/(((1+(t^7)))^3)) with respect to t

Solution

Going through a very long sequence of steps (somewhat shortened by doing severalbasic derivatives at once),

({{({{(d/(dt))}})((((1+sin(3t)))^4)/(((1+(t^7)))^3))=(((((1+(t^7)))^3)(d/(dt))(((1+sin(3t)))^4)-(((1+sin(3t)))^4)(d/(dt))(((1+(t^7)))^3))/((((((1+(t^7)))^3)))^2))=}, {(((((1+(t^7)))^3)4(((1+sin(3t)))^3)3cos(3t)-(((1+sin(3t)))^4)3(((1+(t^7)))^2)7(t^6))/(((1+(t^7)))^6))=}, {(((1+(t^7))4(((1+sin(3t)))^3)3cos(3t)-(((1+sin(3t)))^4)3×7(t^6))/(((1+(t^7)))^4))=}, {((12(1+(t^7))(((1+sin(3t)))^3)cos(3t)-21(((1+sin(3t)))^4)(t^6))/(((1+(t^7)))^4))}})

Note that in practice you might pull the common factor of (((1+sin(3t)))^3) out of the numerator - for example that would be helpful if you had to take the derivative, set it equal to 0, and solve for t.

Problem 34

Find the derivative of (sin^2)((x^3))+(cos^2)((x^3)) with respect to x

Solution

Brute-forcing this we have

({{(d/(dx))(({{(sin^2)((x^3))+(cos^2)((x^3)))}})=2(sin^1)((x^3))(d/(dx))sin((x^3))+2(cos^1)((x^3))(d/(dx))cos((x^3))=}, {2sin((x^3))cos((x^3))(d/(dx))(x^3)+2 cos((x^3))(-sin((x^3))(d/(dx))(x^3))=}, {2sin((x^3))cos((x^3))×3(x^2)-2cos((x^3))sin((x^3))×3(x^2)=}, {6(x^2)sin((x^3))cos((x^3))-6(x^2)sin((x^3))cos((x^3))=0}})

A simpler (but higher-tech) approach is to realize that for any angle θ, (sin^2)(θ)+(cos^2)(θ)=1.  So this problem is really (d/(dx))1=0.

Problem 35

As mentioned above we won’t define the function ln(x) this semester. Assuming you knew nothing about this function other than that (d/(dx))ln|x|=(1/x), what is the derivative of  ln|sec(x)+tan(x)|? (simplify your answer)

Solution

Starting with the chain rule we have

({{({{(d/(dx))}})ln|sec(x)+tan(x)|=(1/(sec(x)+tan(x)))({{(d/(dx))}})(sec(x)+tan(x))=}, {(1/(sec(x)+tan(x)))(sec(x)tan(x)+(sec^2)(x))=(1/(sec(x)+tan(x)))(sec(x)+tan(x))sec(x)=}, {1×sec(x)=sec(x)}})

Problem 36

The equation y=((x-5)^(1/3))+1 has a tangent line at x=5.  What is the derivative y' at x=5?

Solution

Taking the derivative we have

({{({{((dy)/(dx))=(d/(dx))( (((x-5))^(1/3))+1)=(1/3)(((x-5))^(-2/3))(d/(dx))(x-5)+0=}, {(1/3)(((x-5))^(-2/3))×1=(1/(3(((x-5))^(2/3)))).}})}})

The derivative at x=5 is therefore undefined.  However (Underscript[lim , x->5])y'=∞ (you can check this using one-sided limits), which means in terms of the graph there is a vertical tangent line.

Problem 37

Find the equation of the tangent line to the curve ((sin(x))/x)=-((2cos(y))/y) at the point (π,(π/2)).

Solution

First it’s worth checking that the point is actually on the curve itself:

({{({{((sin(π))/π)= ?=((-2cos((π/2)))/(π/2))}, {(0/π)= ? =((-2×0)/(π/2))}, {0=0  ✓}})}})

To find the slope ((dy)/(dx)) at the point in question we need to use implicit differentiation as this equation cannot be solved for y.  So we take the derivative of both sides treating y as a stand-in for a formula in x and using the chain and other rules to introduce ((dy)/(dx)).

({{({{ ((sin(x))/x)=-((2cos(y))/y)}, {(d/(dx)) ((sin(x))/x)=(d/(dx))-((2cos(y))/y)}, {((x(d/(dx))sin(x)-sin(x)(d/(dx))x)/(x^2))=-(((y(d/(dx))(2 cos(y))-2cos(y)(d/(dx))y)/(y^2)))}, {((x cos(x)-sin(x)×1)/(x^2))=-(((y(-2sin(y)((dy)/(dx)))-2cos(y)((dy)/(dx)))/(y^2)))}, {((x cos(x)-sin(x))/(x^2))=((2y sin(y)((dy)/(dx))+2cos(y)((dy)/(dx)) )/(y^2))=((dy)/(dx))(((2y sin(y)+2 cos(y))/(y^2)))}, {((dy)/(dx))=(((x cos(x)-sin(x))/(x^2)))(((y^2)/(2y sin(y)+2 cos(y))))}})}})

Substituting in x=π and y=(π/2) yields

({{({{(Subscript[m, tan])=(((x cos(x)-sin(x))/(x^2)))(((y^2)/(2y sin(y)+2 cos(y))))|({{ }, {x=π,y=π/2}})}})}, {(Subscript[m, tan])=(((π cos(π)-sin(π))/(π^2)))(((((π/2))^2)/(2(π/2) sin(π/2)+2 cos(π/2))))}, {(Subscript[m, tan])=((π(-1)-0)/(π^2))×(((π^2)/4)/(π×1+2×0))=-(1/4)}})

Note: Given that we had to evaluate the tangent slope at a single spot it might have been easier to simply plug in x=π and y=(π/2) before solving for ((dy)/(dx)).  However if you have to find the tangent line at more than one point finding the general formula for ((dy)/(dx)) (as we did here) will save you a lot a time.

The line through (π,(π/2)) with slope -(1/4) is y-(π/2)=-(1/4)(x-π), or y=-(1/4)x+((3π)/4).  You can see the graph of the curve together with the line below

ContourPlot[{ Sin[x]/x==-2Cos[y]/y, y==-1/4 x + 3Pi/4}, {x,0,2Pi}, {y,-Pi,Pi}, AspectRatio->Automatic, Epilog-> {PointSize[Large], Point[ {Pi, Pi/2}]}, PlotPoints-> 100, PlotLegends->Map[ Text[TraditionalForm[#]]&, {Sin[x]/x==-2Cos[y]/y, y==-x/4 + 3Pi/4}], Frame->False, Axes-> True, Ticks-> {Range[0,2Pi, Pi/4], Range[-Pi,Pi,Pi/4]}]

Problem 38

Find the equation of the tangent line to (y^3)-x (y^2)=2 at the point ((3/2),2).

Solution

A quick check will confirm that ((3/2),2) is on the given curve (as (((2))^3)-((3/2))(((2))^2)=8-6=2).  All we need to find is the tangent slope - that is, ((dy)/(dx)) at the point in question.  Differentiating both sides of the equation yields:

({{({{(y^3)-x (y^2)=2}, {(d/(dx))((y^3)-x (y^2))=(d/(dx))2}, {3(y^2)(d/(dx))y-(x (d/(dx))(y^2)+(y^2)(d/(dx))x)=0}, {3(y^2)((dy)/(dx))-(x×2y((dy)/(dx))+(y^2)×1)=0}, {3(y^2)((dy)/(dx))-2x y((dy)/(dx))-(y^2)=0}, {3(y^2)((dy)/(dx))-2x y((dy)/(dx))=(y^2)}, {((dy)/(dx))(3(y^2)-2x y)=(y^2)}, {((dy)/(dx))=((y^2)/(3(y^2)-2x y))=(y/(3y-2x))}})}})

Evaluating this at the point ((3/2),2) gives

({{({{(Subscript[m, tan])=((dy)/(dx))|({{ }, {x=(3/2), y=2}})=(2/(3(2)-2(3/2)))=(2/(6-3))=(2/3)}})}})

So the tangent line to the graph is y-2=(2/3)(x-(3/2)), or y=(2/3)x+1.  Checking this against the graph looks like this:

ContourPlot[ {y^3-x y^2==2,y==2/3x+1}, {x,-2,3}, {y,-1, 4}, AspectRatio-> Automatic, Frame->False, Axes->True, PlotLegends-> Map[ Text[TraditionalForm[#]]&, {y^3-x y^2==2, y==2/3 x + 1}], Epilog-> {PointSize[Large], Point[ {3/2,2}]}, Ticks-> {Range[-10,10,1],Range[-10,10,1]}]

Problem 39

Find ((dy)/(dx)) if tan((x/y))=x-sin(y).

Solution

In this case we “simply” have to differentiate both sides with respect to x and solve for ((dy)/(dx)):

({{({{tan((x/y))=x-sin(y)}, {(d/(dx))tan((x/y))=(d/(dx))(x-sin(y))}, {(sec^2)((x/y))(d/(dx))((x/y))=1-cos(y)(d/(dx))y}, {(sec^2)((x/y))(((y×1-x×(d/(dx))y)/(y^2)))=1-cos(y)((dy)/(dx))}, {(sec^2)((x/y))(((y-x((dy)/(dx)))/(y^2)))=1-cos(y)((dy)/(dx))}, {(sec^2)((x/y))(y-x((dy)/(dx)))=(y^2)(1-cos(y)((dy)/(dx)))}, {y (sec^2)((x/y))-x (sec^2)((y/x))((dy)/(dx))=(y^2)-(y^2)cos(y)((dy)/(dx))}, {(y^2)cos(y)((dy)/(dx))-x (sec^2)((x/y))((dy)/(dx))=(y^2)-y (sec^2)((x/y))}, {((dy)/(dx))((y^2)cos(y)-x (sec^2)((x/y)))=(y^2)-y (sec^2)((x/y))}, {((dy)/(dx))=(((y^2)-y (sec^2)((x/y)))/((y^2)cos(y)-x (sec^2)((x/y))))}})}})

Problem 40

Two sets of curves are called orthogonal if at any point where a curve from the first set intersects a curve from the second set, they intersect at right angles (which would imply their derivatives are “negative reciprocals”).  Verify that the family of circles given by (x^2)+(y^2)=c and the family of lines y=m x are orthogonal.

Solution

Suppose that (x,y) is a point of intersection for the general circle (x^2)+(y^2)=c and the general line y=m x.  Then for the tangent slope on the circle we have:

({{({{(x^2)+(y^2)=c}, {(d/(dx))((x^2)+(y^2))=(d/(dx))c}, {2x+2y((dy)/(dx))=0}, {2y((dy)/(dx))=-2x}, {((dy)/(dx))=((-2x)/(2y))=-(x/y)}})}})

For the line y=m x we have ((dy)/(dx))=m=(y/x) (as the circle’s tangent slope formula uses both y and x it makes sense to convert the line’s slope formula into y and x as well).  The test for orthoginality is that the are “negative reciprocals”, or that the two slopes multiply to -1.  And in fact (Subscript[m, circle])×(Subscript[m, line])=(-(x/y))×((y/x))=1.

Problem 41

Find all points on the ellipse ((x^2)/25)+((y^2)/9)=1 for which the slope of the tangent line is 2.

Solution

First we need to find the slope of the tangent line at any point on the ellipse, which is best done via implicit differentiation:

({{({{((x^2)/25)+((y^2)/9)=1}, {(d/(dx))(((x^2)/25)+((y^2)/9))=(d/(dx))1}, {((2x)/25)+((2y)/9)((dy)/(dx))=0}, {((2y)/9)((dy)/(dx))=-((2x)/25)}, {((dy)/(dx))=(-((2x)/25))/(((2y)/9))=-((2x)/25)×(9/(2y))=-((9x)/(25y))}})}})

As we want a point on the ellipse (so ((x^2)/25)+((y^2)/9)=1) where the tangent slope is 2 (so ((dy)/(dx))=2), we have a system of 2 equations in 2 unknowns: ((x^2)/25)+((y^2)/9)=1 and -((9x)/(25y))=2.  Rewriting the second equation as y=-((9x)/50) and substituting that into the first equation we get:

({{({{((x^2)/25)+((y^2)/9)=1}, {((x^2)/25)+((((-((9x)/50)))^2)/9)=1}, {((x^2)/25)+((9(x^2))/2500)=1}, {100(x^2)+9(x^2)=2500}, {109(x^2)=2500}, {(x^2)=(2500/109)}, {x=±(Sqrt[(2500/109)])=±(50/(Sqrt[109]))}, {⇓}, {((x^2)/25)+((y^2)/9)=1 and  x=±(50/(Sqrt[109]))}, {((((±(50/(Sqrt[109]))))^2)/25)+((y^2)/9)=1}, {((2500/109)/25)+((y^2)/9)=1}, {(100/109)+((y^2)/9)=1}, {900+109(y^2)=981}, {109(y^2)=981-900=81}, {(y^2)=(81/109)}, {y=±(Sqrt[(81/109)])=±(9/(Sqrt[109]))}})}})

At first glance it appears as if we have 4 points (2 choices for x and 2 choices for y), but the equation -((9x)/(25y))=2 implies that x and y have opposite signs.  So the two places on the ellipse where the tangent slope is 2 are ((50/(Sqrt[109])),-(9/(Sqrt[109]))) and (-(50/(Sqrt[109])),(9/(Sqrt[109]))).  Checking these on the graph looks like this:

ContourPlot[ {x^2/25+y^2/9==1, y+9/Sqrt[109]==2(x-50/Sqrt[109]), y-9/Sqrt[109]==2(x+50 /Sqrt[109])}, {x,-7,7},{y,-7,7}, Axes->True, Frame->False, PlotLegends-> Map[ Text[TraditionalForm[#]]&,{x^2/25+y^2/9==1, y+9/Sqrt[109]==2(x-50/Sqrt[109]), y-9/Sqrt[109]==2(x+50 /Sqrt[109])}], Epilog-> {PointSize[Large], Point[{50,-9}/Sqrt[109]], Point[ {-50,9}/Sqrt[109]]}]

Problem 42

Find the second derivative of y=(x^3)cos(3x)

Solution

Finding the derivatives one at a time:

({{({{y'=}})(d/(dx))((x^3)cos(3x))=(x^3)(d/(dx))cos(3x)+cos(3x)(d/(dx))(x^3)=-3(x^3)sin(3x)+3(x^2)cos(3x)}, {
y''=(d/(dx))y'=(d/(dx))(-3(x^3)sin(3x)+3(x^2)cos(3x))=}, {(-3(x^3)(d/(dx))sin(3x)+sin(3x)(d/(dx))(-3(x^3)))+(3(x^2)(d/(dx))cos(3x)+cos(3x)(d/(dx))3(x^2))=}, {-3(x^3)cos(3x)×3+sin(3x)(-9(x^2))+3(x^2)(-sin(3x)×3)+cos(3x)×6x=}, {-9(x^3)cos(3x)-18(x^2)sin(3x)+6x cos(3x)}})

Problem 43

Find the first 20 derivatives of y=3(x^5)-15x+5

Solution

Finding the derivatives one at a time:

({{({{y'=15(x^4)-15}, {y''=60(x^3)}, {y'''=180(x^2)}, {(y^((4)))=360x}, {(y^((5)))=360}, {(y^((6)))=0}, {(y^((7)))=0}, {⋮}, {(y^((20)))=0}})}})

In fact if p(x) is a polynomial of degree n, (p^((k)))(x)=0 whenever k>n.

Problem 44

Find the (312^th) derivative of sin(x)

Solution

Computing the first several derivatives a pattern quickly appears:

Row[ {TableForm[ Table[{Text[k], TraditionalForm[Text[D[Sin[x],{x,k}]]]}, {k,0,11} ], TableHeadings->{None, Text/@{"Derivative", "Formula Number"}}], "   ", TableForm[ Table[{Text[k], TraditionalForm[Text[D[Sin[x],{x,k}]]]}, {k,12,23} ], TableHeadings->{None, Text/@{"Derivative", "Formula Number"}}]}]

Note that the formulas go around in a “loop” of length 4.  Therefore all the (("multiple of 4")^th) derivatives (including the (312^th) derivative) will be sin(x).

Problem 45

Find the third derivative of y=((6(x^2)+1)/(x-3)).

Solution

Computing the derivatives by brute force:

({{({{y'=(((x-3)(d/(dx))(6(x^2)+1)-(6(x^2)+1)(d/(dx))(x-3))/(((x-3))^2))}})=(((x-3)(12x)-(6(x^2)+1)×1)/(((x-3))^2))=((12(x^2)-36x-(6(x^2)+1))/(((x-3))^2))=((6(x^2)-36x-1)/(((x-3))^2))}, { 
y''=(d/(dx))y'=(d/(dx))((6(x^2)-36x-1)/(((x-3))^2))=(((((x-3))^2)(d/(dx))(6(x^2)-36x-1)-(6(x^2)-36x-1)(d/(dx))(((x-3))^2))/((((((x-3))^2)))^2))=}, {(((((x-3))^2)(12x-36)-(6(x^2)-36x-1)×2(x-3)(d/(dx))(x-3))/(((x-3))^4))=(((((x-3))^2)(12x-36)-(6(x^2)-36x-1)×2(x-3)×1)/(((x-3))^4))=}, {(((x-3)(12x-36)-2(6(x^2)-36x-1))/(((x-3))^3))=(((12(x^2)-36x-36x+108)-(12(x^2)-72x-2))/(((x-3))^3))=(110/(((x-3))^3))}, { }, {y'''=(d/(dx))y''=(d/(dx))(110/(((x-3))^3))=(d/(dx))110(((x-3))^(-3))=110(-3)(((x-3))^(-4))(d/(dx))(x-3)=}, {-330(((x-3))^(-4))×1=-(330/(((x-3))^4))}})

If you are wondering why the form for the second derivative collapsed so nicely try doing the long division in ((6(x^2)+1)/(x-3)) before you take the derivative.  You will find that ((6(x^2)+1)/(x-3))=6x+18+(55/(x-3)); the sudden simplification is due to the fact that the second derivative of the 6x+18 part is 0.

Problem 46

If cos(x- y)=x+y, find y'' as a function of x and y

Solution

The trick here is to find y' implicity, then take the derivative of that (which will introduce new copies of y', which you will have a formula for already):

({{({{cos(x- y)=x+y}, {(d/(dx))cos(x -y)=(d/(dx))(x+y)}, {-sin(x-y)(d/(dx))(x-y)=(d/(dx))x+(d/(dx))y}, {-sin(x -y)((d/(dx))x-(d/(dx))y)=1+((dy)/(dx))}, {-sin(x -y)(1-((dy)/(dx)))=1+((dy)/(dx))}, {- sin(x- y)+ sin(x- y)((dy)/(dx))=1+((dy)/(dx))}, {- sin(x- y)-1=((dy)/(dx))- sin(x- y)((dy)/(dx))}, {- sin(x- y)-1=((dy)/(dx))(1- sin(x- y))}, {((dy)/(dx))=((-sin(x- y)-1)/(1- sin( x- y)))=((1+sin(x-y))/(sin(x-y)-1))}})}})

Taking another derivative gives us

({{({{y''=(d/(dx))y'=(d/(dx))}})((1+sin(x-y))/(sin(x-y)-1))=}, {(((sin(x-y)-1)(d/(dx))(1+sin(x-y))-(1+sin(x-y))(d/(dx))(sin(x-y)-1))/(((sin(x-y)-1))^2))=(((sin(x-y)-1)cos(x-y)(d/(dx))(x-y)-(1+sin(x-y))cos(x-y)(d/(dx))(x-y))/(((sin(x-y)-1))^2))=}, {(((sin(x-y)-1)cos(x-y)(1-((dy)/(dx)))-(1+sin(x-y))cos(x-y)(1-((dy)/(dx))))/(((sin(x-y)-1))^2))=}, {(((sin(x-y)-1)cos(x-y)(1-((1+sin(x-y))/(sin(x-y)-1)))-(1+sin(x-y))cos(x-y)(1-((1+sin(x-y))/(sin(x-y)-1))))/(((sin(x-y)-1))^2))}})

If you are ambitious enough to clear denominators and simplify some nice agebra will happen, with the end result being y''=((4cos(x-y))/(((sin(x-y)-1))^3)).  I leave it for the “interested student to show”.

Problem 47

Find all vertically-opening parabolas (that is, quadratics of the form y=a (x^2)+b x +c) which pass through (1,2), have a tangent slope of 3 at (1,2), and whose second derivative at (1,2) is 4.

Solution

What we have been given is essentially y(1)=2, y'(1)=3,and y''(1)=4.  Using the formula y=a (x^2)+b x +c we have y'=2 a x + b and y''=2a.   Therefore y(1)=a×(1^2)+b×1+c=a+b+c, y'(1)=2a(1)+b=2a+b, and y''(1)=2a.  Setting the various formulas at 1 equal to each other gives the system of equations a+b+c=2, 2a+b=3, and 2a=4.  Solving the last equation gives a=2; using that value in the second equation gives 4+b=3, or b=-1; and using both values in the first equation gives 2+(-1)+c=2, or c=1.  So the only quadratic that first the 3 requirements is y=2(x^2)-x+1.

Problem 48

Find all values of λ for which y=sin(λ x) is a solution to y''+4y=0.

Solution

If y=sin(λ x) then y'=cos(λ x)×λ=λ cos(λ x) and y''=(d/(dx))λ cos(λ x)=λ(d/(dx)) cos(λ x)=λ(-λ sin(λ x))=-(λ^2)sin(λ x).  So for y''+4y=0 we would have to have (-(λ^2)sin(λ x))+4(sin(λ x))=0, or (4-(λ^2))sin(λ x)=0.  This canbe true for all x only if 4-(λ^2)=0, which is the same as (λ^2)=4, or λ=±(Sqrt[4])=±2.  So the only functions of the form sin(λ x) which are solutions to the equations are sin(2x) and sin(-2x).

Problem 49

Explain why that if all you have is the graph of y=f(x) (and not is formula or other information), you can still estimate the graph of y=f'(x).

Solution

As f'(x) measures the slope of the tangent line to y=f(x) at each point, you can take a discrete set of x-values, estimate the slope of the tangent line at each one, and come up with a set of new points of the form (x-value, estimated tangent slope at that x-value).  These points will be on the graph of y=f'(x), so playing the “connect the dots” will give you a decent estimate for the graph.

Problem 50

Below is the graph of some function y=f(x).  Estimate the graph of y=f'(x).

Show[ Graphics[{Opacity[.3], Table[ InfiniteLine[ {{0,k},{1,k}}], {k,-10,10,.5}], Table[ InfiniteLine[ {{k,0},{k,1}}], {k,-10,10,.5}]}],Plot[ {(x^3-4x^2)/2^x},{x,-2,8}], AspectRatio-> Automatic, Axes-> True, PlotRange-> {{-4,8},{-7,3}}, Ticks-> {Range[-10,10,.5], Range[-10,10,.5]}]

Solution

Using the idea of Problem 50 we can estimate the slope of the tangent lines at various x-values from -1 to 8 and use these to generate a new graph.  The grid makes it easier to estimate the slopes of tangent lines using an index card or another piece of paper - slide the card/paper into position so its edge is a tangent line  and then find some convenient grid points to estimate the slope as ((Δ y)/(Δ x)).  When you do so you get a table like this (I did a lot of values spaced .5 units apart in x - your estimates may differ slightly):

TableForm[Table[ {x,Round[(2^(-x)) (-8 x+3 (x^2))-(2^(-x)) (-4 (x^2)+(x^3)) Log[2],.1]}, {x,0,8,.5}], TableHeadings-> {None, Text/@ {"x", "estimated slope at x"}}]

Graphing these points and playng “connect the dots” will give you a decent graph of the derivative:

Plot[ (2^(-x)) (-8 x+3 (x^2))-(2^(-x)) (-4 (x^2)+(x^3)) Log[2], {x,0,8}, Epilog-> {PointSize[Large],Point[Table[ {x,Round[(2^(-x)) (-8 x+3 (x^2))-(2^(-x)) (-4 (x^2)+(x^3)) Log[2],.05]}, {x,0,8,.5}]]}]

Problem 51

Below is the graph of some function y=f(x).  Estimate the graph of y=f'(x).

Show[ Graphics[{Opacity[.3], Table[ InfiniteLine[ {{0,k},{1,k}}], {k,-10,10,1/2}], Table[ InfiniteLine[ {{k,0},{k,1}}], {k,-10,10,1/2}]}],Plot[ {1.3Sin[Pi x]Cos[Pi/2 x]},{x,0,6}], AspectRatio-> Automatic, Axes-> True, PlotRange-> {{-2,8},{-3,3}}, Ticks-> {Range[-10,10,.5], Range[-10,10,.5]}]

Solution

Playing the same game as in Problem 50 we can come up with some estimated points on the graph of f'(x):

TableForm[Table[ {x,Round[4.084070449666731` Cos[((π x)/2)] Cos[π x]-2.0420352248333655` Sin[((π x)/2)] Sin[π x],.1]}, {x,0,6,.5}], TableHeadings-> {None, Text/@ {"x", "estimated slope at x"}}]

Note that a lot of these estimates (and yours may differ slightly) are the same in absolute value - this is due to the symmetries in the curve.

Placing these in a new graph and playing “connect the dots” in a nice smooth way (so different people will have slightly different graphs - you could always estimate and plot more points if needed) gives a good estimate of the graph of y=f'(x):

Plot[ 4.084070449666731` Cos[((π x)/2)] Cos[π x]-2.0420352248333655` Sin[((π x)/2)] Sin[π x], {x,0,6}, Epilog-> {PointSize[Large],Point[Table[ {x,Round[4.084070449666731` Cos[((π x)/2)] Cos[π x]-2.0420352248333655` Sin[((π x)/2)] Sin[π x],.05]}, {x,0,6,.5}]]}]