Chain Rule Explained With Examples (Step by Step)

If $y = f(g(x))$, then:

$$\frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) \cdot g'(x)$$

In words: differentiate the outside function, leave the inside alone, then multiply by the derivative of the inside.

In Leibniz notation, if $y$ depends on $u$ and $u$ depends on $x$:

$$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

Why it works: rates multiply. If $u$ changes 3 times as fast as $x$, and $y$ changes 2 times as fast as $u$, then $y$ changes $2 \cdot 3 = 6$ times as fast as $x$.

Ask: "What would I calculate last if I plugged in a number?" That last operation is the outside function.

Function	Outside	Inside
$(3x + 1)^5$	$u^5$	$3x + 1$
$\sin(x^2)$	$\sin u$	$x^2$
$e^{4x}$	$e^u$	$4x$
$\ln(x^2 + 1)$	$\ln u$	$x^2 + 1$
$\sqrt{1 - x^2}$	$\sqrt{u}$	$1 - x^2$

If the inside is just plain $x$, the inside derivative is 1 and you do not need the chain rule.

Example: Differentiate $y = (2x^3 - 5)^4$.

1. Outside: $u^4$. Inside: $u = 2x^3 - 5$.
2. Differentiate the outside and keep the inside unchanged: $4(2x^3 - 5)^3$.
3. Differentiate the inside: $6x^2$.
4. Multiply: $4(2x^3 - 5)^3 \cdot 6x^2 = 24x^2(2x^3 - 5)^3$.

Using Leibniz notation for the same problem: $\frac{dy}{du} = 4u^3$ and $\frac{du}{dx} = 6x^2$, so $\frac{dy}{dx} = 4u^3 \cdot 6x^2 = 24x^2(2x^3 - 5)^3$.

Example: Differentiate $\sin(5x^2)$.

1. Outside: $\sin u$, whose derivative is $\cos u$. Inside: $5x^2$.
2. Outside derivative with the inside kept: $\cos(5x^2)$.
3. Inside derivative: $10x$.
4. Answer: $10x\cos(5x^2)$.

Example: Differentiate $\cos^3 x$.

1. Rewrite it so the structure is clear: $\cos^3 x = (\cos x)^3$. The outside is the cube, and the inside is $\cos x$.
2. Outside derivative: $3(\cos x)^2$.
3. Inside derivative: $-\sin x$.
4. Answer: $-3\sin x \cos^2 x$.

Notice the difference: in $\sin(5x^2)$ the trig function is outside, while in $\cos^3 x$ the power is outside.

$$\frac{d}{dx}\left[e^{u}\right] = e^{u} \cdot u' \qquad \frac{d}{dx}\left[\ln u\right] = \frac{u'}{u}$$

Example: Differentiate $e^{x^2}$.

1. Outside: $e^u$, which is its own derivative. Inside: $x^2$.
2. Answer: $e^{x^2} \cdot 2x = 2xe^{x^2}$.

Example: Differentiate $\ln(x^2 + 1)$.

1. Outside: $\ln u$, with derivative $\frac{1}{u}$. Inside: $x^2 + 1$.
2. Answer: $\frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}$.

Some functions have three or more layers. Peel them one at a time, from the outside in, and multiply every derivative together.

Example: Differentiate $\sin^3(2x)$.

1. Rewrite: $\left[\sin(2x)\right]^3$. The layers are: cube, then sine, then $2x$.
2. Derivative of the cube: $3\sin^2(2x)$.
3. Derivative of the sine: $\cos(2x)$.
4. Derivative of $2x$: $2$.
5. Multiply: $3\sin^2(2x) \cdot \cos(2x) \cdot 2 = 6\sin^2(2x)\cos(2x)$.

Example: Differentiate $h(x) = x^2(3x + 1)^4$.

1. The whole thing is a product, so start with the product rule: $f = x^2$ and $g = (3x + 1)^4$.
2. $f' = 2x$.
3. For $g'$, use the chain rule: $4(3x + 1)^3 \cdot 3 = 12(3x + 1)^3$.
4. Product rule: $h'(x) = 2x(3x + 1)^4 + x^2 \cdot 12(3x + 1)^3$.
5. Factor out $2x(3x + 1)^3$: $h'(x) = 2x(3x + 1)^3\left[(3x + 1) + 6x\right]$.
6. Simplify: $h'(x) = 2x(3x + 1)^3(9x + 1)$.

Paste a function into the VeilGPT solver to check each layer of your chain rule work.