What is the difference between Chain Rule and Power Rule?

Power Rule says \frac{d}{dx}x^n = nx^{n-1}, which works when the base is just x. Chain Rule is more general: it gives \frac{d}{dx}(g(x))^n = n(g(x))^{n-1} \cdot g'(x) when the base is any function g(x). If you apply Power Rule to x^n and then compare it to Chain Rule on (g(x))^n, you see that Chain Rule multiplies by g'(x) at the end—that is the extra step.

Do I need to use Chain Rule every time I see a composite function?

Chain Rule is the formal technique for differentiating composites. Sometimes a composite can be rewritten in simpler form before differentiating. For example, (2x)^3 = 8x^3, which you can differentiate with Power Rule alone. But Chain Rule is always correct and often the most direct route, especially when expansion is messy or impossible.

Can I apply Chain Rule multiple times in the same problem?

Yes. If you differentiate \sin(x^2 + 1), you apply Chain Rule once. But if you differentiate \sin((2x + 1)^3), you apply Chain Rule on the outside (for the sine), then again on the inside (for the power), because the argument of sine is itself composite. Work from the outside in, applying Chain Rule to each layer until you reach x.

What if the inner function is very complicated?

Chain Rule works exactly the same way. Identify outer and inner, differentiate each, and multiply. The difficulty lies only in computing g'(x), which might require Power Rule, Product Rule, Quotient Rule, or other techniques. The structure of Chain Rule itself never changes—only the complexity of the pieces.

Chain Rule

The Chain Rule tells you how to differentiate a composite function—when one function is plugged into another—by multiplying the outer and inner derivatives.

\frac{d}{dx}f\big(g(x)\big) = f'\big(g(x)\big)\,g'(x)

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What each symbol means

What Chain Rule takes

Symbol	Meaning
$f$	The outer function—the one applied last in the composition $f(g(x))$ . It takes the output of $g$ as its input.
$g$	The inner function—the one applied first in the composition $f(g(x))$ . Its output becomes the input to $f$ .
$x$	The independent variable, the thing you are ultimately differentiating with respect to; confusing it with the intermediate variable $u = g(x)$ is a common source of error.

When to use it

Use Chain Rule when you see one function nested inside another, like $\sin(x^2)$ or $(3x+1)^5$ .

Level

Usually taught in: Calculus I · Appears on: AP Calculus

Worked examples

1. Differentiate a polynomial raised to a power

Problem

Find

\frac{d}{dx}(4x + 3)^6

$\text{Outer function: } f(u) = u^6, \quad \text{Inner function: } g(x) = 4x + 3$
Identify which part is the outer function (the last operation) and which is the inner (applied first). Here, we raise something to the 6th power (outer), and that something is $4x + 3$ (inner).
$f'(u) = 6u^5$
Differentiate the outer function with respect to its input $u$ .
$g'(x) = 4$
Differentiate the inner function with respect to $x$ .
$\frac{d}{dx}(4x + 3)^6 = f'(g(x)) \cdot g'(x) = 6(4x + 3)^5 \cdot 4 = 24(4x + 3)^5$
Apply Chain Rule: the derivative of the outer function evaluated at $g(x)$ , multiplied by the derivative of the inner function.

Answer: $24(4x + 3)^5$

We cannot use Power Rule directly because the base $4x + 3$ is not just $x$ . Chain Rule handles this by reducing the exponent (getting $6$ in front and an exponent of $5$ ), keeping the inner expression, and multiplying by the derivative of that inner expression, which is $4$ .

2. Differentiate a trigonometric function with a polynomial argument

Problem

Find

\frac{d}{dx}\cos(x^2 - 3x)

$\text{Outer function: } f(u) = \cos(u), \quad \text{Inner function: } g(x) = x^2 - 3x$
The cosine is applied to an expression that depends on $x$ , so cosine is the outer function and $x^2 - 3x$ is the inner.
$f'(u) = -\sin(u)$
The derivative of cosine is negative sine, found using the standard derivative rule.
$f'(g(x)) = -\sin(x^2 - 3x)$
Evaluate the outer derivative at the inner function $g(x) = x^2 - 3x$ .
$g'(x) = 2x - 3$
Differentiate the inner function $x^2 - 3x$ with respect to $x$ using Power Rule.
$\frac{d}{dx}\cos(x^2 - 3x) = -\sin(x^2 - 3x) \cdot (2x - 3)$
Multiply the derivative of the outer function (evaluated at the inner function) by the derivative of the inner function.

Answer: $-\sin(x^2 - 3x) \cdot (2x - 3)$

Without Chain Rule, we would have no way to handle trigonometric functions of complicated arguments—none of the basic derivative rules apply directly. Chain Rule breaks this into two familiar pieces: the derivative of cosine and the derivative of the polynomial inside it.

3. Find the rate of change of a garden plot's area

Problem

A rectangular garden plot is being enlarged. Its side length grows as

s(t) = \sqrt{t}

meters at time

t

seconds, and its area is

A(s) = s^2

square meters. Find

\frac{dA}{dt}

when

t = 4

$\text{Outer function: } A(s) = s^2, \quad \text{Inner function: } s(t) = \sqrt{t} = t^{1/2}$
The area depends on the side length $s$ , which depends on time $t$ . Area is the outer function; the side length is the inner.
$\frac{dA}{ds} = 2s$
Differentiate area with respect to side length using Power Rule.
$\frac{ds}{dt} = \frac{1}{2}t^{-1/2} = \frac{1}{2\sqrt{t}}$
Differentiate the side length with respect to time, again using Power Rule with the exponent $\frac{1}{2}$ .
$\frac{dA}{dt} = \frac{dA}{ds} \cdot \frac{ds}{dt} = 2s \cdot \frac{1}{2\sqrt{t}}$
Apply Chain Rule in Leibniz notation: how area changes with time equals how area changes with side length, times how side length changes with time.
$\text{Substitute } s = \sqrt{t}: \quad \frac{dA}{dt} = 2\sqrt{t} \cdot \frac{1}{2\sqrt{t}} = \frac{2\sqrt{t}}{2\sqrt{t}} = 1$
Use $s = \sqrt{t}$ from the given function to simplify. The result is constant.
$\text{At } t = 4: \quad \frac{dA}{dt} = 1 \text{ m}^2/\text{s}$
Even without substitution, the answer is always $1$ square meter per second, regardless of the time.

Answer: $\frac{dA}{dt} = 1 \text{ m}^2/\text{s}$

This is a real-world scenario where quantity you care about (area) depends on a measurement (side length) that itself changes over time. Chain Rule connects these rates: it tells how fast the area grows with time, given how fast the side length grows with time. This kind of related-rates problem is one of the most important applications of Chain Rule.

Common mistakes

Where Chain Rule usually goes wrong

The mistake	Why it is wrong	The fix
For $\frac{d}{dx}\sin(3x)$ , writing the answer as $\cos(3x)$ and stopping.	The outer derivative is $\cos(3x)$ , but Chain Rule requires you to multiply by the derivative of the inner function, which is $3$ .	$\frac{d}{dx}\sin(3x) = \cos(3x) \cdot 3 = 3\cos(3x)$
For $\frac{d}{dx}(5x - 2)^7$ , writing the answer as $7(5x - 2)^6$ without multiplying by $5$ .	You applied the power rule correctly to the outside, but Chain Rule says you must also multiply by the derivative of what is inside the parentheses, which is $5$ .	$\frac{d}{dx}(5x - 2)^7 = 7(5x - 2)^6 \cdot 5 = 35(5x - 2)^6$
For $\frac{d}{dx}\ln(x^2 - 1)$ , writing the answer as $\frac{1}{x^2 - 1}$ and forgetting the rest.	The derivative of $\ln(u)$ is $\frac{1}{u}$ , which gives $\frac{1}{x^2 - 1}$ . But Chain Rule requires multiplying by the derivative of $x^2 - 1$ , which is $2x$ .	$\frac{d}{dx}\ln(x^2 - 1) = \frac{1}{x^2 - 1} \cdot 2x = \frac{2x}{x^2 - 1}$

Tips and when to use something else

Always identify the outer and inner functions before you start differentiating—draw a box around the innermost part if it helps.
Forgetting to multiply by the inner derivative is the most common Chain Rule mistake. After writing $f'(g(x))$ , always ask yourself: have I multiplied by $g'(x)$ ?
If the function is not composite, do not force Chain Rule. For example, $\frac{d}{dx}x^3 = 3x^2$ by Power Rule alone; there is no need to invoke Chain Rule.
When functions are nested inside functions inside functions—like $\sin(e^{x^2})$ —apply Chain Rule repeatedly, working from the outside in. Each application handles one layer.

Related formulas

Power RuleThe Power Rule states that the derivative of x^n is nx^{n-1}, making it the essential tool for differentiating any polynomial expression.Product RuleProduct Rule lets you find the derivative of a product of two functions without multiplying them out first; use it when the product is complex.Derivative of Sin xThe derivative of sin x is cos x, giving the slope of the sine curve; use this rule when differentiating any sine function.Derivative of Cos xThe derivative of cosine x equals negative sine x, the fundamental rule for finding rates of change of oscillating cosine functions.Implicit DifferentiationUse implicit differentiation to find dy/dx when an equation defines y implicitly instead of as an explicit function of x.

Frequently asked questions

What is the difference between Chain Rule and Power Rule?: Power Rule says $\frac{d}{dx}x^n = nx^{n-1}$ , which works when the base is just $x$ . Chain Rule is more general: it gives $\frac{d}{dx}(g(x))^n = n(g(x))^{n-1} \cdot g'(x)$ when the base is any function $g(x)$ . If you apply Power Rule to $x^n$ and then compare it to Chain Rule on $(g(x))^n$ , you see that Chain Rule multiplies by $g'(x)$ at the end—that is the extra step.
Do I need to use Chain Rule every time I see a composite function?: Chain Rule is the formal technique for differentiating composites. Sometimes a composite can be rewritten in simpler form before differentiating. For example, $(2x)^3 = 8x^3$ , which you can differentiate with Power Rule alone. But Chain Rule is always correct and often the most direct route, especially when expansion is messy or impossible.
Can I apply Chain Rule multiple times in the same problem?: Yes. If you differentiate $\sin(x^2 + 1)$ , you apply Chain Rule once. But if you differentiate $\sin((2x + 1)^3)$ , you apply Chain Rule on the outside (for the sine), then again on the inside (for the power), because the argument of sine is itself composite. Work from the outside in, applying Chain Rule to each layer until you reach $x$ .
What if the inner function is very complicated?: Chain Rule works exactly the same way. Identify outer and inner, differentiate each, and multiply. The difficulty lies only in computing $g'(x)$ , which might require Power Rule, Product Rule, Quotient Rule, or other techniques. The structure of Chain Rule itself never changes—only the complexity of the pieces.

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Reviewed 2026-09-18