18.01 Solutions, Problem Set 2

• §3.1 #1 Find the derivative of each function:

(a) 6x^{9};

Solution: \frac{d}{dx}(6x^{9}) = 6 · 9x^{9−1} = 54x^{8}.

(b) π^{5};

Solution: \frac{d}{dx}(π^{5}) = 0 since π^{5} is a constant.

(g) x^{4} + x^{3} + x^{2} + x + 1. Solution:

\begin{matrix}\frac{d}{dx}(x^{4} + x^{3} + x^{2} + x + 1) \\ = \frac{d}{dx}(x^{4}) + \frac{d}{dx}(x^{3}) + \frac{d}{dx}(x^{2}) + \frac{d}{dx}(x) + \frac{d}{dx}(1) \\ = 4x^{3} + 3x^{2} + 2x + 1 \end{matrix} .

• §3.1 #3b If s = 13 − 9t + 6t^{3} is the position at time t of an object moving on a straight line, find the velocity v and the acceleration a.

Solution: The velocity is

v = \frac{ds}{dt} = \frac{d}{dt}(13 − 9t + 6t^{3}) = −9 + 18t^{2}.

The acceleration is

a = \frac{dv}{dt} = \frac{d}{dt}(−9 + 18t^{2}) = 36t.

• §3.1 #5 Find the line tangent to the curve y = 3x^{2} − 5x + 2 at the point (2, 4).

Solution: Evaluating \frac{dy}{dx} = \frac{d}{dx}(3x^{2} − 5x + 2) = 6x − 5 at x = 2 shows the slope of the tangent line is 6 · 2 − 5 = 7. The equation of the tangent line is therefore

y − 4 = 7(x − 2), or y = 7x − 10.

• §3.1 #6 Find the points on the curve y = 4x^{3} +6x^{2} −24x+10 at which the tangent line is horizontal.

Solution: The slope of the tangent line is

\frac{dy}{dx} = \frac{d}{dx}(4x^{3} + 6x^{2} − 24x + 10) = 12x^{2} + 12x − 24.

The tangent line is horizontal when the slope is zero, so we solve 12x^{2} + 12x − 24 = 0. Factoring

12x^{2} + 12x − 24 = 12(x^{2} + x − 2) = 12(x + 2)(x − 1),

we find x = −2 or x = 1. The points are therefore (−2, 50) and (1, −4).

• §3.1 #18

Find the tangent to the curve y = x^{3} that passes through the point (0, 2).

Solution: The derivative of y = x^{3} is \frac{d}{dx}(x^{3}) = 3x^{2}, so the slope fo the

tangent line at the point (a, a^{3}) is 3a^{2}. The equation of the tangent line is therefore y − a^{3} = 3a^{2}(x − a). For this to pass through the point (0, 2), we must have

\begin{aligned}2 − a^{3} & = & 3a^{2}(−a); \\ 2 − a^{3} & = & −3a^{3}; \\ 2 & = & −2a^{3}; \\ −1 & = & a^{3}. \end{aligned}

So a = −1 and the equation of the tangent line is

y + 1 = 3(x + 1), or y = 3x + 2.

• §3.2 #4

Differentiate (x − 1)(x^{4} + x^{3} + x^{2} + x + 1) in two ways and verify that your answers agree.

Solution: If we first simplify

(x−1)(x^{4}+x^{3}+x^{2}+x+1) = (x^{5}+x^{4}+x^{3}+x^{2}+x)−(x^{4}+x^{3}+x^{2}+x+1) = x^{5}−1,

then we get

\frac{d}{dx}[(x − 1)(x^{4} + x^{3} + x^{2} + x + 1)] = \frac{d}{dx}(x^{5} − 1) = 5x^{4}.

On the other hand, by the product rule

\frac{d}{dx}[(x − 1)(x^{4} + x^{3} + x^{2} + x + 1)]

= (x − 1)\frac{d}{dx}(x^{4} + x^{3} + x^{2} + x + 1) + (x^{4} + x^{3} + x^{2} + x + 1)\frac{d}{dx}(x − 1)

= (x − 1)(4x^{3} + 3x^{2} + 2x + 1) + (x^{4} + x^{3} + x^{2} + x + 1) · 1

= (4x^{4} + 3x^{3} + 2x^{2} + x) − (4x^{3} + 3x^{2} + 2x + 1) + (x^{4} + x^{3} + x^{2} + x + 1) = 5x^{4}.

• §3.2 Differentiate each function and simplify your answer as much as possible.

9. \frac{x + 1}{x − 1}.

Solution: By the quotient rule,

\frac{d}{dx} (\frac{x + 1}{x − 1}) = \frac{(x − 1)\frac{d}{dx}(x + 1) − (x + 1)\frac{d}{dx}(x − 1)}{(x − 1)^{2}}
= \frac{(x − 1) − (x + 1)}{(x − 1)^{2}}
= \frac{−2}{(x − 1)^{2}} .
10. \frac{1}{x^{2} + 2}

Solution:

\begin{aligned}\frac{d}{dx} (\frac{1}{x^{2} + 2}) & = & \frac{(x^{2} + 2)\frac{d}{dx}(1) − 1 · \frac{d}{dx}(x^{2} + 2)}{(x^{2} + 2)^{2}} \\ & = & \frac{−2x}{(x^{2} + 2)^{2}} . \end{aligned}

14. \frac{4x − x^{4}}{x^{3} + 2} .

Solution:

\frac{d}{dx} (\frac{4x − x^{4}}{x^{3} + 2} ) = \frac{(x^{3} + 2)(4 − 4x^{3}) − (4x − x^{4})(3x^{2})}{(x^{3} + 2)^{2}}
= \frac{−4x^{6} − 4x^{3} + 8 + 3x^{6} − 12x^{3}}{(x^{3} + 2)^{2}}
= \frac{−x^{6} − 16x^{3} + 8}{(x^{3} + 2)^{2}}.

28. x^{4} − \frac{1}{x^{2} − 1}.

Solution:

\begin{aligned}\frac{d}{dx} (x^{4} − \frac{1}{x^{2} − 1}) & = & 4x^{3} − \frac{−\frac{d}{dx}(x^{2} − 1)}{(x^{2} − 1)^{2}} \\ & = & 4x^{3} + \frac{2x}{(x^{2} − 1)^{2}} . \end{aligned}

• §3.2 #40

Find the tangent to y = \frac{x^{3} + x}{x − 1} at (2, 10).

Solution: By the quotient rule,

\begin{aligned}\frac{d}{dx} (\frac{x^{3} + x}{x − 1}) & = & \frac{(x − 1)(3x^{2} + 1) − (1)(x^{3} + x)}{(x − 1)^{2}} \\ & = & \frac{2x^{3} − 3x^{2} − 1}{(x − 1)^{2}}. \end{aligned}

Evaluating this at x = 2 gives 3, so the equation is

y − 10 = 3(x − 2), or y = 3x + 4.

• §3.3

Find dy/dx, if

2. y = (4 + 5x)^{4}; Solution: By the power rule (and chain rule),

\frac{d}{dx}(4 + 5x)^{4} = 4(4 + 5x)^{3} \frac{d}{dx}(4 + 5x) = 20(4 + 5x)^{3}.

7. y = \frac{1}{(3x + 1)^{4}} ;

Solution:

\frac{d}{dx}(3x + 1)^{−4} = −4(3x + 1)^{−5} \frac{d}{dx}(3x + 1) = \frac{−12}{(3x + 1)^{5}} .

20. y = (1 − 3x)^{−1}; Solution:

\frac{d}{dx}(1 − 3x)^{−1} = −(1 − 3x)^{−2}(−3) = 3(1 − 3x)^{−2}.

• §3.3 #43 Find the equation of the tangent line to the curve y = (x^{3} − x^{2} + x)^{8} at the point (1, 1).

Solution: The derivative is

\frac{dy}{dx} = \frac{d}{dx}(x^{3} − x^{2} + x)^{8} = 8(x^{3} − x^{2} + x)^{7}(3x^{2} − 2x + 1).

Evaluating at x = 1 gives 16, so the equation is

y − 1 = 16(x − 1) or y = 16x − 15.

• §8.3 Find the derivative dy/dx of the given function.

2. y = \frac{1}{2}(e^{x} − e^{−x}).

Solution: The derivative is

\frac{dy}{dx} = \frac{d}{dx} \frac{1}{2}(e^{x} − e^{−x}) = \frac{1}{2} (\frac{d}{dx}(e^{x}) − \frac{d}{dx}(e^{−x}))
= \frac{1}{2} (e^{x} − (e^{−x})\frac{d}{dx}(−x))
= \frac{1}{2}(e^{x} + e^{−x}).

5. y = e^{e^{x}}.

Solution: Applying the chain rule with u = e^{x} gives

\frac{dy}{dx} = \frac{d}{dx}(e^{e^{x}})e^{e^{x}} \frac{d}{dx}(e^{x}) = e^{e^{x}}e^{x} = e^{x+e^{x}}.

6. y = x^{e} + e^{x}.

Solution: The power rule with exponent e gives

\frac{dy}{dx} = \frac{d}{dx}(x^{e}) + \frac{d}{dx}(e^{x}) = ex^{e−1} + e^{x}.

8. y = (3x + 1)e^{−3x}. Solution: The product rule gives

\begin{aligned}\frac{dy}{dx} & = & (3x + 1)\frac{d}{dx}(e^{−3x}) + (e^{−3x})\frac{d}{dx}(3x + 1) \\ & = & (−3)(3x + 1)e^{−3x} + 3e^{−3x} = −9xe^{−3x}. \end{aligned}

• §8.4 #2 Find dy/dx in each case:

(a) y = ln(3x + 2); Solution: The chain rule gives

\frac{d}{dx} ln(3x + 2) = \frac{1}{3x + 2} · \frac{d}{dx}(3x + 2) = \frac{3}{3x + 2}.

(e) y = x ln x − x; Solution: The product rule gives

\frac{d}{dx}(x ln x) − \frac{d}{dx}(x) = x\frac{d}{dx} ln x + ln x − 1 = x · \frac{1}{x} + ln x − 1 = ln x.

(j) y = ln(ln(x)).

Solution: The chain rule gives

\frac{d}{dx} ln(ln x) = \frac{1}{ln x} · \frac{d}{dx}(ln x) = \frac{1}{x ln x}.

• §3.4 Find dy/dx.

2. y = cos(x^{5} + 1).

Solution: By the chain rule,

\frac{d}{dx} cos(x^{5} + 1) = − sin(x^{5} + 1)\frac{d}{dx}(x^{5} + 1) = −5x^{4} sin(x^{5} + 1).

13. y = tan(sin x).

Solution: Since \frac{d}{dx}(tan u) = sec^{2} u\frac{du}{dx}, we have

\frac{d}{dx} tan(sin x) = sec^{2}(sin x)\frac{d}{dx}(sin x) = (cos x) sec^{2}(sin x).

25. y = \frac{cos x}{x}.

Solution: By the quotient rule

\frac{dy}{dx} = \frac{x\frac{d}{dx}(cos x) − cos x\frac{d}{dx}x}{x^{2}} = \frac{−x sin x − cos x}{x^{2}}.

• §3.4 #34 Find the values of x for which the graph of y = x+2 sin x has a horizontal tangent.

Solution: Setting the derivative

\frac{d}{dx}(x + 2 sin x) = 1 + 2 cos x

equal to zero and solving for x gives

\begin{aligned}1 + 2 cos x & = & 0 \\ cos x & = & −1/2. \end{aligned}

This happens when x = ±2π/3 (and we can add to this any positive or negative multiple of 2π).

• §3.4 #36 By differentiating the first of the following double angle formulas, obtain the second:

sin(2x) = 2 sin x cos x, cos(2x) = cos^{2} x − sin^{2} x.

Solution: Differentiating both sides of the first formula gives

\frac{d}{dx} sin(2x) = \frac{d}{dx}2 sin x cos x 2 cos(2x) = 2[sin x\frac{d}{dx} cos x + cos x\frac{d}{dx} sin x] 2 cos(2x) = 2(− sin^{2} x + cos^{2} x) .

Halving both sides of this equation gives the second formula.

• §9.2 Find the derivative dy/dx of the given function

15. y = e^{2x} sin(3x). Solution: The product rule (followed by chain rule) gives \frac{dy}{dx} = e^{2x} \frac{d}{dx} sin(3x) + sin(3x)\frac{d}{dx}e^{2x} = 3e^{2x} cos(3x) + 2e^{2x} sin(3x).

16. y = sin(ln x^{2}). Solution: The chain rule applied twice gives

\frac{dy}{dx} = \frac{d}{dx}(sin(2 ln x)) = cos(2 ln x)\frac{d}{dx}(2 ln x) = 2x^{−1} cos(2 ln x) .