4.3 The chain rule

For parts (a) and (b), remember the definition of a derivative:

When $h$ is positive, $U+h$ is an increased urchin population; what is the sign of $K(U+h)-K(U)$?

For part (c), use the chain rule!

(a) $\diff{K}{U}$ is negative (b) $\diff{U}{O}$ is negative (c) $\diff{K}{O}$ is positive

Remember that Leibniz notation suggests fractional cancellation.

negative

If $g(x)=\cos x$ and $h(x)=5x+3$, then $f(x)=g(h(x))$. So we apply the chain rule, with “outside" function $\cos x$ and “inside" function $5x+3$.

$-5\sin(5x+3)$

You can expand this into a polynomial, but it's easier to use the chain rule. If $g(x)=x^5$, and $h(x)=x^2+2$, then $f(x)=g(h(x))$.

$10x(x^2+2)^4$

You can expand this into a polynomial, but it's easier to use the chain rule. If $g(k)=k^{17}$, and $h(k)=4k^4+2k^2+1$, then $T(k)=g(h(k))$.

$17(4k^4+2k^2+1)^{16}\cdot(16k^3+4k)$

If we define $g(x)=\sqrt{x}$ and $h(x)=\dfrac{x^2+1}{x^2-1}$, then $f(x)=g(h(x))$.
To differentiate the square root function: $\ds\diff{}{x}\{\sqrt{x}\}=\ds\diff{}{x}\left\{x^{1/2}\right\}=\dfrac{1}{2}x^{-1/2}=\dfrac{1}{2\sqrt{x}}$.

$\frac{-2x}{({x^2-1})\sqrt{x^4-1}}$

You'll need to use the chain rule twice.

$-e^{\cos(x^2)}\cdot \sin(x^2)\cdot 2x$

Use the chain rule.

$-4$

Use the chain rule.

$[\cos x -x\sin x]e^{x\cos(x)}$

Use the chain rule.

$[2x-\sin x]e^{x^2+\cos(x)}$

Use the chain rule.

$\frac{3}{2\sqrt{x-1}\sqrt{x+2}^3}$

Recall $\dfrac{1}{x^2}=x^{-2}$ and $\sqrt{x^2-1}=(x^2-1)^{1/2}$.

$f'(x)= -\dfrac{2}{x^3}+\dfrac{x}{\sqrt{x^2-1}}$ is defined for $x$ in $(-\infty,1) \cup (1,\infty)$.

$f'(x)=\dfrac{(1+x^2)(5\cos 5x)-(\sin 5x)(2x)}{{(1+x^2)}^2}$

If we let $g(x)=\sec x$ and ${h(x)}=e^{2x+7}$, then $f(x)=g(h(x))$, so by the chain rule, $f'(x)=g'(h(x))\cdot h'(x)$. However, in order to evaluate $h'(x)$, we'll need to use the chain rule again.

$2e^{2x+7}\sec(e^{2x+7})\tan(e^{2x+7})$

What trig identity can you use to simplify the first factor in the equation?

$y=1$

Velocity is the derivative of position with respect to time. In this case, the velocity of the particle is given by $s'(t)$.

$t=\frac{2}{3}$ and $t=4$

The slope of the tangent line is the derivative.
You'll need to use the chain rule twice.

$2e\sec^2(e)$

Start with the product rule, then use the chain rule to differentiate $e^{4x}$.

$y'=4e^{4x}\tan x+e^{4x}\sec^2 x$

Start with the quotient rule; you'll need the chain rule only to differentiate $e^{3x}$.

$\dfrac{3}{{(1+e^{3})}^2}$

More than one chain rule needed here.

$2 \sin(x) \cdot \cos(x) \cdot e^{\sin^2(x)}$

More than one chain rule application is needed here.

$\cos\left(e^{5x}\right)\cdot e^{5x}\cdot 5$

More than one chain rule application is needed here.

$-e^{\cos(x^2)}\cdot \sin(x^2) \cdot 2x$

More than one chain rule application is needed here.

$y'=-\sin\big(x^2+\sqrt{x^2+1}\big)\left(2x+\dfrac{x}{\sqrt{x^2+1}}\right)$

What rule do you need, besides chain? Also, remember that $\cos^2x = [\cos x]^2$.

$y'=2x\cos^2x-2(1+x^2) \sin x\cos x$

$y'=\dfrac{e^{3x}(3x^2-2x+3)}{(1+x^2)^2}$

$-40$

The product of two functions is zero exactly when at least one of the functions is zero.

$(1,1)$ and $(-1,-1)$.

If $t \ge 1$, then $0<\frac{1}{t} \leq 1$.

Always

The notation $\cos^3(5x-7)$ means $\left[\cos(5x-7)\right]^3$. So, if $g(x)=x^3$ and $h(x)=\cos(5x-7)$, then $g(h(x))=\left[\cos(5x+7)\right]^3=\cos^3(5x+7)$.

$e^x\sec^3(5x-7)(1+15\tan(5x-7))$

In Example 4.1.11, we generalized the product rule to three factors:

This isn't strictly necessary, but it will simplify your computations.

$e^{2x} \cos 4x + 2x e^{2x} \cos 4x -4 x e^{2x} \sin 4x$

At time $t$, the particle is at the point $\big(x(t),y(t)\big)$, with $x(t)=\cos t$ and $y(t)=\sin t$. Over time, the particle traces out a curve; let's call that curve $y=f(x)$. Then $y(t) = f\big(x(t)\big)$, so the slope of the curve at the point $\big(x(t),y(t)\big)$ is $f'\big(x(t)\big)$. You are to determine the values of $t$ for which $f'\big(x(t)\big)=-1$.

$t=\dfrac{\pi}{4}$

Set $f(x) = e^{x+x^2}$ and $g(x)=1+x$. Compare $f(0)$ and $g(0)$, and compare $f'(x)$ and $g'(x)$.

Let $f(x)=e^{x+x^2}$ and $g(x)=1+x$. Then $f(0)=g(0)=1$.

$f'(x)=(1+2x)e^{x+x^2}$ and $g'(x)=1$. When $x>0$,

Since $f(0)=g(0)$, and $f'(x)>g'(x)$ for all $x>0$, that means $f$ and $g$ start at the same place, but $f$ always grows faster. Therefore, $f(x)>g(x)$ for all $x>0$.

If $\sin 2x$ and $2\sin x \cos x$ are the same, then they also have the same derivatives.

$\cos(2x)=\cos^2x-\sin^2x$

This is a long, nasty problem, but it doesn't use anything you haven't seen before. Be methodical, and break the question into as many parts as you have to. At the end, be proud of yourself for your problem-solving abilities and tenaciousness!

To sketch the curve, you can start by plotting points. Alternately, consider $x^2+y$.

(a)

The particle traces the curve $y=1-x^2$ restricted to domain $[-1,1]$. At $t=0$, the particle is at the top of the curve, $(1,0)$. Then it moves to the right, and goes back and forth along the curve, repeating its path every $2\pi$ units of time.
(b) $\sqrt{3}$

Use a similar method to Question 32, Section 4.3.

(a) $f'(x)=(1+2x)e^{x+x^2}$ $f''(x)=(4x^2+4x+3)e^{x+x^2}$ $h'(x)=1+3x$ $h''(x)=3$
(b) $f(0)=h(0)=1$; $f'(0)=h'(0)=1$; $f''(0)=h''(0)=3$
(c) $f$ and $h$ “start at the same place," since $f(0)=h(0)$. Also $f'(0)=h'(0)$, and $f''(x)=(4x^2+4x+3)e^{x+x^2} > 3e^{x+x^2}>3=h''(x)$ when $x>0$. Since $f'(0)=h'(0)$, and since $f'$ grows faster than $h'$ for positive $x$, we conclude $f'(x)>h'(x)$ for all positive $x$. Now we can conclude that (since $f(0)=h(0)$ and $f$ grows faster than $h$ when $x>0$) also $f(x)>h(x)$ for all positive $x$.