AP Calculus BC study materials
AP Calculus BC study resources

AP Calculus BC: Differentiation: Composite, Implicit, and Inverse Functions

AP Calculus BC · Work through your exam, one topic at a time.

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Differentiation: Composite, Implicit, and Inverse Functions

Core ideas

The chain rule is the engine of this domain: $\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$. Layer functions from the outside in and track each inner derivative. This extends to inverse functions via $\left(f^{-1}\right)'(b)=\frac{1}{f'(a)}$ where $f(a)=b$, and to inverse trigonometric derivatives.

Implicit differentiation handles relations not solved for $y$: differentiate both sides with respect to $x$, treating $y$ as a function of $x$ and attaching $\frac{dy}{dx}$ whenever you differentiate a $y$-term. Then solve algebraically for $\frac{dy}{dx}$. Logarithmic differentiation is a useful technique when a function has a variable in both base and exponent.

Traps

  • Forgetting the inner derivative in the chain rule, especially with trig, exponential, and inverse trig compositions.
  • In implicit differentiation, dropping the $\frac{dy}{dx}$ factor on $y$-terms or on products involving $y$.
  • For inverse functions, mixing up which point's derivative belongs in the denominator; the reciprocal relationship pairs the input of one with the output of the other.
  • The derivative of $\arctan x$ is $\frac{1}{1+x^2}$, not $\frac{1}{\sqrt{1-x^2}}$ (that is arcsine).

Worked example

Find $\frac{dy}{dx}$ for $x^2+y^2=25$. Differentiate: $2x+2y\frac{dy}{dx}=0$, so $\frac{dy}{dx}=-\frac{x}{y}$. At the point $(3,4)$, the slope is $-\frac{3}{4}$, matching the tangent to the circle.