Practicing the chain rule

Question

I'm trying to understand the answer to the problem in this screenshot

The goal is to use the chain rule to find the derivative of $$e^{\sin(x^2)}$$ As seen in the screenshot, the correct answer is $$2x e^{\sin(x^2)}\cos(x^2)$$

From my understanding the chain goes:

$$f'(g'(h(x))) h'(x) g'(x) $$

but I am not too sure how it gets $\cos(x^2) $

Answer 1

Your statement of the chain rule is incorrect. The derivative of $f(g(h(x)))$ with respect to $x$ is $f'(g(h(x)))g'(h(x))h'(x)$.

In your case, $f(x) = e^x$, $g(x) = \sin(x)$, and $h(x) = x^2$, so $$f(g(h(x))) = f(g(x^2)) = f(\sin(x^2)) = e^{\sin(x^2)}$$

To apply the chain rule, we need the derivatives of $f$, $g$, and $h$, which are: $$\begin{aligned} f'(x) &= e^x \\ g'(x) &= \cos(x) \\ h'(x) &= 2x \\ \end{aligned}$$ Then the three factors of the chain rule are: $$\begin{aligned} f'(g(h(x)) &= f'(\sin(x^2)) = e^{\sin(x^2)} \\ g'(h(x)) &= \cos(x^2) \\ h'(x) &= 2x \\ \end{aligned} $$ Multiplying these together, we get the final answer: $$f'(g(h(x)))g'(h(x))h'(x) = 2x e^{\sin(x^2)}\cos(x^2)$$ which matches the correct answer at your link.

Answer 2

Let me supplement @Bungo's answer by explaining why the derivative of $f(g(h(x)))$ with respect to $x$ is $$ f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \, . $$ Recall that the chain rule states that (under certain conditions) $$ \frac{d}{dx}\left(u(v(x))\right)=u'(v(x)) \cdot v'(x) $$ Let $u=f$ and $v=g\circ h$.* Then, $$ \frac{d}{dx}\left(u(v(x))\right)=f'(g(h(x)))\cdot\frac{d}{dx}\left(g(h(x))\right) $$ But we know that $$ \frac{d}{dx}\left(g(h(x))\right)=g'(h(x)) \cdot h'(x) \, . $$ Hence, overall $$ \frac{d}{dx}\left(u(v(x))\right)=\frac{d}{dx}(f(g(h(x))))=f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \, . $$ In cleaner notation, $$ (u \circ v)'(x) = (f \circ g \circ h)'(x) = (f' \circ g \circ h)(x) \cdot (g' \circ h)(x) \cdot h'(x) \, . $$

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*The notation $v=g\circ h$ just means that $v(a)=g(h(a))$ for all $a$.

Answer 3

I think rather than memorizing rules, a simpler approach is to just use differentials and substitution. Let's start with the formula:

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

Now, math doesn't care about the variables we use. So, we can simply say $u = \sin(x^2)$, which makes the formula:

$$e^u$$

If we differentiate this (not the full derivative, just the differential), we get:

$$d(e^u) = e^u\,du$$

This is a 100% true statement! But what is $du$? Well, we can find that by just differentiating both sides of $u = \sin(x^2)$. However, to do that, we need another substitution! Let's set $w = x^2$, so $u = \sin(w)$. That means that we can find $du$ by saying:

$$ du = \cos(w)\,dw $$

So what is $dw$? Well, $w = x^2$, and we can differentiate that directly:

$$ dw = 2x\,dx $$

Now we just build it all back up.

$$ e^u\, du = e^{\sin(x^2)}\,\cos(w)\,dw \\ = e^{\sin(x^2)}\,\cos(x^2)\,2x\,dx $$

To get the derivative, just divide by $dx$:

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

The rule works just fine, but doing it this way is super-helpful for me for recognizing (a) what the rule is and (b) why the rule is.

Answer 4

Let $f: z\mapsto \mathrm e^z\\g:y\mapsto \sin y\\h: x\mapsto x^2$

So $f': z\mapsto \mathrm e^z\\g':y\mapsto \cos y\\h': x\mapsto 2x$

Then we may apply the chain rule, by whatever route you prefer: $$\begin{align}(\mathrm e^{\sin(x^2)})'&=[(f\circ g\circ h)'](x) &&=\dfrac{\mathrm df(g(h(x)))}{\mathrm d x} \\[1ex]&=[(g\circ h)'\cdot(f'\circ g\circ h)](x) &&= \dfrac{\mathrm d g(h(x))}{\mathrm d x}\cdot\dfrac{\mathrm d f(g(h(x))}{\mathrm d g(h(x))} \\[1ex]&=[h'\cdot(g'\circ h)\cdot(f'\circ g\circ h)](x)&& =\dfrac{\mathrm d h(x)}{\mathrm d x}\cdot \dfrac{\mathrm d g(h(x))}{\mathrm d h(x)}\cdot\dfrac{\mathrm d f(g(h(x))}{\mathrm d g(h(x))} \\[1ex]&= h'(x)\cdot g'(h(x))\cdot f'(g(h(x))) &&=\dfrac{\mathrm d x^2}{\mathrm d x}\cdot \dfrac{\mathrm d \sin(x^2)}{\mathrm d x^2}\cdot\dfrac{\mathrm d \mathrm e^{\sin(x^2)}}{\mathrm d \sin(x^2)}\\[1ex]&=(x^2)'\cdot\sin'(x^2)\cdot\mathrm e^{\sin(x^2)}&&=2x\cdot\cos(x^2)\cdot\mathrm e^{\sin(x^2)}\end{align}$$