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

(a) More urchins means less kelp, and fewer urchins means more kelp. This means kelp and urchins are negatively correlated, so $\diff{K}{U}<0$.

If you aren't sure why that is, we give a more detailed explanation here, using the definition of the derivative. When $h$ is a positive number, $U+h$ is greater than $U$, so $K(U+h)$ is less than $U$, hence $K(U+h)-K(U)<0$. Therefore:

Similarly, when $h$ is negative, $U+h$ is less than $U$, so $K(U+h)-K(U)>0$, and

Therefore:

(b) More otters means fewer urchins, and fewer otters means more urchins. So, otters and urchins are negatively correlated: $\diff{U}{O}<0$.
(c) Using the chain rule, $\diff{K}{O} = \diff{K}{U}\cdot\diff{U}{O}$. Parts (a) and (b) tell us both these derivatives are negative, so their product is positive: $\diff{K}{O}>0$.

We can also see that $\diff{K}{O}>0$ by thinking about the relationships as described. When the otter population increases, the urchin population decreases, so the kelp population increases. That means when the otter population increases, the kelp population also increases, so kelp and otters are positively correlated. The chain rule is a formal version of this kind of reasoning.

Remember that Leibniz notation suggests fractional cancellation.

negative

since we multiply three positive quantities and one 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 $5x+3$.

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

Applying the chain rule:

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 $f(x)=g(h(x))$.

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

Using the chain rule,

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 $T(k)=g(h(k))$.

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

Using the chain rule,

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: $\ds\diff{}{x}\{\sqrt{x}\}=\ds\diff{}{x}\left\{x^{1/2}\right\}=\dfrac{1}{2}x^{-1/2}=\dfrac{1}{2\sqrt{x}}$.

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

Using the chain rule:

And now, the quotient rule:

You'll need to use the chain rule twice.

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

If we let $g(x)=e^x$ and $h(x)=\cos(x^2)$, then $f(x)=g(h(x))$, so $f'(x)=g'(h(x))\cdot h'(x)$.

In order to evaluate $\diff{}{x}\{{\cos(x^2)}\}$, we'll need the chain rule again.

Use the chain rule.

$-4$

We use the chain rule, followed by the quotient rule:

When $x=2$:

Use the chain rule.

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

Using the chain rule, followed by the product rule:

Use the chain rule.

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

Using the chain rule:

Use the chain rule.

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

Using the chain rule, followed by the quotient rule:

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 $(-\infty,1) \cup (1,\infty)$.

First, we manipulate our function to make it easier to differentiate:

Now, we can use the power rule to differentiate $\dfrac{1}{x^2}$. This will be easier than differentiating $\dfrac{1}{x^2}$ using quotient rule, but if you prefer, quotient rule will also work.

The function $f(x)$ is only defined when $x \neq 0$ and when $x^2-1 \geq 0$. That is, when $x$ is in $(-\infty,-1] \cup [1,\infty)$. We have an added restriction on the domain of $f'(x)$: $x^2-1$ must not be zero. So, the domain of $f'(x)$ is $(-\infty,-1)\cup(1,\infty)$.

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

We use the quotient rule, noting that $\ds\diff{}{x}\{\sin 5x\}=5\cos 5x$:

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.

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

If we let $g(x)=\sec x$ and $h(x)=e^{2x+7}$, then $f(x)=g(h(x))$, so by the chain rule, $f'(x)=g'(h(x))\cdot h'(x)$. Since $g'(x)=\sec x \tan x$:

Here, we need the chain rule again:

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

$y=1$

It is possible to start in on this problem with the product rule and then the chain rule, but it's easier if we simplify first. Since $\tan^2x+1=\sec^2 x=\frac{1}{\cos^2x}$, we see

for all values of $x$ for which $\cos x$ is nonzero. That is, $f(x)=1$ for every $x$ that is not an integer multiple of $\pi/2$ (and $f(x)$ is not defined when $x$ is an integer multiple of $\pi/2$). Therefore, $f'(x)=0$ for every $x$ on which $f$ exists, and in particular $f'(\pi/4)=0$. Also, $f(\pi/4)=1$, so the tangent line to $f$ at $x=\pi/4$ is the line with slope 0, passing through the point $(\pi/4,1)$:

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 $t=4$

Velocity is the derivative of position with respect to time. So, the velocity of the particle is given by $s'(t)$. We need to find $s'(t)$, and determine when it is zero.

To differentiate, we us the chain rule.

To determine where this function is zero, we factor:

So, the velocity is zero when $e^{{t^3-7t^2+8t}}=0$, when $3t-2=0$, and when $t-4=0$. Since $e^{{t^3-7t^2+8t}}$ is never zero, this tells us that the velocity is zero precisely when $t=\frac{2}{3}$ or $t=4$.

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

$2e\sec^2(e)$

The slope of the tangent line is the derivative. If we let $f(x)=\tan x$ and $g(x)=e^{x^2}$, then $f(g(x))=\tan(e^{x^2})$, so $y'=f'(g(x)) \cdot g'(x)$:

We find ourselves once more in need of the chain rule:

Finally, we evaluate this derivative at the point $x=1$:

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

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

Using the Product rule,

and the chain rule:

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

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

Using the quotient rule,

Now, the chain rule:

So, when $x=1$:

More than one chain rule needed here.

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

This requires us to apply the chain rule twice.

More than one chain rule application is needed here.

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

This requires us to apply the chain rule twice.

More than one chain rule application is needed here.

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

We'll use the chain rule twice.

More than one chain rule application is needed here.

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

We start with the chain rule:

and find ourselves in need of chain rule a second time:

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

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

Using the product rule,

Here, we'll need to use the chain rule. Remember $\cos^2 x = [\cos x]^2$.

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

We use the quotient rule, noting by the chain rule that $\diff{}{x}\{e^{3x}\}=3e^{3x}$:

$-40$

By the chain rule,

Using the product rules and the result above,

Plugging in $x=2$:

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

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

Let $f(x)=xe^{-(x^2-1)/2}=xe^{(1-x^2)/2}$. Then, using the product rule,

Here, we need the chain rule: