Many real-world functions are compositions — one function applied inside another. For example, $h(x) = \sin(x^2)$ is $\sin$ applied to $x^2$. The chain rule tells us how to differentiate these.
If $h(x) = f(g(x))$, where $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then:
$$h'(x) = f'(g(x)) \cdot g'(x)$$
In words: Derivative of the outer function (evaluated at the inner function) times the derivative of the inner function.
Leibniz form: If $y = f(u)$ and $u = g(x)$, then $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$.
A useful memory device: "Outside, then Inside." Differentiate the outer shell, keep the inside unchanged, then multiply by the derivative of the inside.
Differentiate each function.
Differentiate $f(x) = \cos(5x^3 - 2x)$.
The chain rule applied to $[g(x)]^n$ gives a commonly used special case:
$$\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1} \cdot g'(x)$$
This works for any real exponent $n$ — including fractions and negatives.
Differentiate $f(x) = \sqrt{4 - x^2} = (4-x^2)^{1/2}$.
$$f'(x) = \frac{1}{2}(4-x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{4-x^2}}$$
This gives the slope of the upper semicircle of radius 2 at any point.
When a function requires both the product/quotient rule and the chain rule, apply from the outermost layer inward.
Differentiate $g(x) = x^2 \sin(3x)$.
Product rule: $g'(x) = 2x\sin(3x) + x^2 \cdot \cos(3x)\cdot 3 = 2x\sin(3x) + 3x^2\cos(3x)$
A composite function $f(g(x))$ and its derivative — notice how the chain rule accounts for both layers of change.
Figure 3.1 — Chain Rule: $h(x)=\sin(x^2)$ and $h'(x)=2x\cos(x^2)$
Some curves are defined by equations like $x^2 + y^2 = 25$ that cannot easily be solved for $y$ explicitly. Implicit differentiation lets us find $\frac{dy}{dx}$ without isolating $y$.
Find $\dfrac{dy}{dx}$ and the equation of the tangent line at $(3, 4)$.
Step 1: Differentiate both sides w.r.t. $x$: $$2x + 2y\frac{dy}{dx} = 0$$
Step 2: Solve for $\dfrac{dy}{dx}$: $$\frac{dy}{dx} = -\frac{x}{y}$$
Step 3: At $(3,4)$: slope $= -\dfrac{3}{4}$
Step 4: Tangent line: $y - 4 = -\dfrac{3}{4}(x-3)$, which simplifies to $3x + 4y = 25$.
Use implicit differentiation to find $\dfrac{dy}{dx}$ for $x^3 + y^3 = 6xy$.
Find $\dfrac{dy}{dx}$ for $\sin(xy) = x + y$.
Differentiate left side using chain rule + product rule:
$\cos(xy) \cdot (y + x\dfrac{dy}{dx}) = 1 + \dfrac{dy}{dx}$
Expand: $y\cos(xy) + x\cos(xy)\dfrac{dy}{dx} = 1 + \dfrac{dy}{dx}$
Collect $\dfrac{dy}{dx}$: $x\cos(xy)\dfrac{dy}{dx} - \dfrac{dy}{dx} = 1 - y\cos(xy)$
Solve: $\dfrac{dy}{dx} = \dfrac{1 - y\cos(xy)}{x\cos(xy) - 1}$
AP Exam Tip: On the AP exam, implicit differentiation often appears in FRQ problems asking you to find a tangent line, a normal line, or to verify a given expression for $\dfrac{dy}{dx}$. Always show your differentiation steps clearly — partial credit is awarded for the setup even if arithmetic errors occur later.
The circle $x^2+y^2=25$ with its tangent line at $(3,4)$. Drag the point to see how the slope $-x/y$ changes.
Figure 3.2 — Implicit Curve: Circle with Tangent Line via Implicit Differentiation
We derive these using implicit differentiation. Let $y = \arcsin(x)$, so $\sin(y) = x$. Differentiating: $\cos(y)\dfrac{dy}{dx} = 1$, giving $\dfrac{dy}{dx} = \dfrac{1}{\cos(y)} = \dfrac{1}{\sqrt{1-x^2}}$.
| Function | Derivative | Domain |
|---|---|---|
| $\arcsin(x)$ | $\dfrac{1}{\sqrt{1-x^2}}$ | $|x| < 1$ |
| $\arccos(x)$ | $\dfrac{-1}{\sqrt{1-x^2}}$ | $|x| < 1$ |
| $\arctan(x)$ | $\dfrac{1}{1+x^2}$ | all $x$ |
| $\text{arccot}(x)$ | $\dfrac{-1}{1+x^2}$ | all $x$ |
| $\text{arcsec}(x)$ | $\dfrac{1}{|x|\sqrt{x^2-1}}$ | $|x| > 1$ |
Combined with the chain rule, these become powerful. For instance, $\dfrac{d}{dx}[\arctan(3x)] = \dfrac{1}{1+(3x)^2} \cdot 3 = \dfrac{3}{1+9x^2}$.
Find $\dfrac{d}{dx}\left[\arcsin\!\left(\tfrac{x}{2}\right)\right]$.
Let $u = x/2$. Then $\dfrac{d}{dx}[\arcsin(u)] = \dfrac{1}{\sqrt{1-u^2}} \cdot \dfrac{1}{2} = \dfrac{1}{2\sqrt{1-x^2/4}} = \dfrac{1}{\sqrt{4-x^2}}$
Find $\dfrac{d}{dx}[\arctan(x^2)]$.
AP Exam Tip: You must memorize the derivatives of $\arcsin$, $\arccos$, and $\arctan$ — they appear on nearly every AP Calculus exam, both MC and FRQ. Note that $\dfrac{d}{dx}[\arcsin] + \dfrac{d}{dx}[\arccos] = 0$ — they are negatives of each other.
$\arctan(x)$ and its derivative $\dfrac{1}{1+x^2}$. The derivative is always positive (arctan is always increasing) and approaches 0 as $|x|\to\infty$.
Figure 3.3 — $\arctan(x)$ and its derivative $\dfrac{1}{1+x^2}$
Differentiate $f(x) = (2x^3 - 5)^4$.
Find the derivative of $g(x) = e^{x^2 - 3x}$.
Differentiate $h(x) = \dfrac{1}{(x^2+1)^3}$. Hint: rewrite as $(x^2+1)^{-3}$.
Use implicit differentiation to find $\dfrac{dy}{dx}$ for $x^2 + 3y^2 = 12$.
Find the slope of the tangent to $x^2 - xy + y^2 = 7$ at the point $(1, 3)$.
Differentiate $f(x) = \arctan\!\left(\dfrac{x}{3}\right)$.
Find $\dfrac{d}{dx}[\arcsin(\sqrt{x})]$.
(AP Style) Let $x^2y + y^3 = 10$. Find $\dfrac{dy}{dx}$ and determine whether the curve has a horizontal tangent at the point where $x = 0$.
If $y = f(g(x))$, then $\dfrac{dy}{dx} = f'(g(x)) \cdot g'(x)$. Differentiate the outer function, then multiply by the derivative of the inner function.
Differentiate both sides of an equation with respect to $x$. When differentiating a $y$-term, apply chain rule: $\frac{d}{dx}[y^n] = ny^{n-1}\frac{dy}{dx}$.
Differentiate a geometric/physical relationship with respect to time $t$. Write an equation relating variables, then differentiate implicitly using chain rule.
$(\arcsin u)' = \dfrac{u'}{\sqrt{1-u^2}}$ $(\arctan u)' = \dfrac{u'}{1+u^2}$
$\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$
$\dfrac{d}{dx}[f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$
$\dfrac{dy}{dx} = -\dfrac{F_x}{F_y}$ where $F(x,y) = 0$ is the implicit equation.
$\dfrac{d}{dx}[u^n] = nu^{n-1} \cdot u'$ for any differentiable $u$.