Inside This Unit: The Full Breakdown
The chain rule handles derivatives of composite functions, implicit differentiation works with curves defined by equations in x and y, and inverse function derivatives round out the differentiation toolkit.
Why it matters
The chain rule appears in nearly every calculus problem from this point forward. In BC specifically, it is essential for parametric derivatives, polar derivatives, and differentiating series term by term.
Key concepts
- Chain rule: d/dx[f(g(x))] = f'(g(x)) * g'(x).
- Implicit differentiation applies the chain rule to y-terms, then solves for dy/dx.
- Inverse function derivative: (f^{-1})'(b) = 1/f'(f^{-1}(b)).
- Inverse trig derivatives: d/dx[arctan x] = 1/(1+x^2), d/dx[arcsin x] = 1/sqrt(1-x^2).
Chain Rule Mastery
The chain rule is used whenever one function is inside another. Differentiate the outer function, keep the inner unchanged, and multiply by the derivative of the inner. For multiply-nested compositions, apply the rule from outside in. For example, d/dx[e^(sin(x^2))] = e^(sin(x^2)) * cos(x^2) * 2x. BC students encounter the chain rule in parametric differentiation (dy/dx = (dy/dt)/(dx/dt)) and when differentiating power series term by term.
Implicit Differentiation
For equations like x^2y + y^3 = 7, differentiate every term with respect to x. Apply the product rule to x^2y and the chain rule to y^3 (giving 3y^2 dy/dx). Collect all dy/dx terms on one side and factor. The result is a formula for dy/dx in terms of both x and y. You can find the second derivative d^2y/dx^2 by differentiating dy/dx implicitly again, substituting the expression for dy/dx wherever it appears.
Inverse Functions and Their Derivatives
If y = f^{-1}(x), then f(y) = x. Differentiating both sides gives f'(y) * dy/dx = 1, so dy/dx = 1/f'(y) = 1/f'(f^{-1}(x)). The AP exam tests this with tables: given values of f and f', find the derivative of f^{-1} at a specified input. For inverse trig functions, the derivatives involve square roots and rational expressions that reappear in integration formulas for arctan and arcsin.
AP exam tip
When differentiating implicitly and the problem asks for d^2y/dx^2, substitute your expression for dy/dx back into the result to express the second derivative entirely in terms of x and y.
Connections to other units
- Unit 4: Related rates require implicit differentiation with respect to time.
- Unit 6: The chain rule in reverse becomes u-substitution for integration.
- Unit 9: Parametric and polar derivatives are built on the chain rule.
