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Calculus I (Differential & intro Integral) Unit 3: Chain rule, implicit & inverse differentiation

These rules let you differentiate functions built inside other functions, curves that aren't solved for y, and inverse functions.

Covers: Chain rule, implicit differentiation, derivatives of inverse & inverse trig functions, higher derivatives

Study guide

The chain rule differentiates compositions: d/dx f(g(x)) = f′(g(x))·g′(x), the derivative of the outside function evaluated at the inside, times the derivative of the inside. For example, d/dx (3x² + 1)⁵ = 5(3x² + 1)⁴·6x. Many functions nest several layers, so apply the rule from the outside in.

Implicit differentiation handles equations where y isn't isolated, like x² + y² = 25. Differentiate both sides with respect to x, treating y as a function of x, so every y-term gets a factor of dy/dx by the chain rule (d/dx y² = 2y·dy/dx). Then solve for dy/dx. The result usually involves both x and y.

Inverse functions have related derivatives: if g = f⁻¹, then g′(a) = 1/f′(g(a)), so the slopes at matching points are reciprocals. This gives the inverse trig derivatives: d/dx arcsin x = 1/√(1 − x²), d/dx arctan x = 1/(1 + x²). It also gives other exponential and log derivatives: d/dx aˣ = aˣ ln a and d/dx logₐ x = 1/(x ln a). The other trig derivatives follow from the quotient rule, such as d/dx tan x = sec²x.

Higher derivatives are derivatives of derivatives. The second derivative f″ measures how the slope changes, and for motion it's acceleration.

On an implicit curve, a horizontal tangent occurs where the numerator of dy/dx is 0 and the denominator is not. Solve that condition together with the curve's original equation to find the actual points, and confirm each point lies on the curve. A vertical tangent occurs where the denominator of dy/dx is 0 and the numerator is not. Equivalently, it occurs where dx/dy = 0. Again, solve together with the original equation and check that the points are on the curve.

Terms that multiply x and y need the product rule: d/dx(xy) = y + x·dy/dx and d/dx(x²y) = 2xy + x²·dy/dx. Collect all dy/dx terms on one side, then factor dy/dx out. To find tangent lines that pass through a point not on the curve, call the unknown point of tangency (a, b). Write the slope two ways, as dy/dx at (a, b) and as the slope between (a, b) and the given point, then solve those together with the curve's equation.

With tables of values, compute h′(a) for h(x) = f(g(x)) as f′(g(a))·g′(a). Look up g(a) first, then find f′ at that output, not at a.

Key ideas

• Chain rule: differentiate the outside, keep the inside, multiply by the derivative of the inside.
• Implicit differentiation: differentiate both sides with respect to x, attach dy/dx to every y-term, then solve for dy/dx.
• Inverse functions: the slope of f⁻¹ at a point is the reciprocal of the slope of f at the matching point.
• Higher derivatives are just repeated differentiation: f″ is the derivative of f′.

Formulas and rules

Common mistakes

• Forgetting the inner derivative in the chain rule.
• In implicit problems, forgetting dy/dx on y² (it becomes 2y·dy/dx).

Words to know

Chain rule

[f(g(x))]′ = f′(g(x))·g′(x)

Implicit differentiation

Differentiate both sides in x, treating y as a function of x: d/dx(y²) = 2y·dy/dx.

Derivative of an inverse function at a

(f⁻¹)′(a) = 1 / f′(f⁻¹(a))

d/dx tan x

sec² x

d/dx arctan x

1/(1 + x²)

d/dx arcsin x

1/√(1 − x²)

d/dx 2ˣ

2ˣ·ln 2

Composite function

f(g(x)): a function inside another.

d/dx aˣ

aˣ ln a

d/dx logₐ x

1/(x ln a)

d/dx sec x

sec x tan x

Higher derivatives

f″, f‴, …: derivatives of derivatives.

Logarithmic differentiation

Take ln of both sides, then differentiate implicitly.

Sample practice questions

41 more questions like these, from level 1 to 5, are in the practice set.