Chain Rule Implicit Differentiation Worksheet Answer Key – AP Calculus AB

Problem 1 Chain rule with a power

Outer: 4(·)³. Inner derivative: 6. So f′(x) = 4(6x + 1)³·6 = 24(6x + 1)³.

Problem 2 Chain rule with a root

Write f(x) = (3x² + 16)^1/2. Then f′(x) = ½(3x² + 16)^−1/2·6x = 3x√(3x² + 16).

Problem 3 Chain rule with e

The derivative of e^u is e^u times u′. So f′(x) = (2x − 3)e^{x2 − 3x}.

Problem 4 Chain rule with a logarithm

The derivative of ln u is u′/u. So f′(x) = 2x + 4x² + 4x.

Problem 5 Chain rule with sine

Outer: cos(3x²). Inner derivative: 6x. So f′(x) = 6x cos(3x²).

Problem 6 Chain rule at a point

f′(x) = 3(2x² − 1)²·4x. At x = 1: 3(1)²·4 = 12.

Problem 7 Chain rule from a table

The chain rule gives h′(1) = f′(g(1))·g′(1) = f′(2)·g′(1) = (−3)(4) = −12.

Problem 8 Product rule with the chain rule

Product rule: 3x²e^−2x + x³·(−2)e^−2x. Factor: f′(x) = e^−2x(3x² − 2x³).

Problem 9 Quotient rule with the chain rule

Quotient rule with (sin 2x)′ = 2 cos 2x: f′(x) = 2 cos 2x·x² − sin 2x·2xx⁴ = 2x cos 2x − 2 sin 2xx³.

Problem 10 Slope on a circle

Differentiate: 2x + 2y dydx = 0, so dydx = −xy. At (5, 12) the slope is −5/12.

Problem 11 Implicit differentiation with a product

Differentiate: 2xy + x² dydx + 3y² dydx = 0. Solve: dydx = −2xyx² + 3y². At (1, 2): −4/(1 + 12) = −4/13.

Problem 12 Tangent line to an implicit curve

Differentiate: 2x + y + xdydx + 2ydydx = 0. So dydx = −(2x + y)x + 2y. At (2, 1): −5/4. The tangent line is y − 1 = −5/4(x − 2), or y = −5/4 x + 7/2.

Problem 13 Derivative of an inverse function

Find the input with f(x) = 10: f(2) = 8 + 2 = 10. Also f′(x) = 3x² + 1, so f′(2) = 13. Then g′(10) = 1/f′(2) = 1/13.

Problem 14 Inverse derivative from a table

Since f(2) = 7, g(7) = 2. So g′(7) = 1/f′(2) = 1/3.

Problem 15 Inverse trigonometric derivatives

(a) The derivative of arctan u is u′/(1 + u²). So the answer is 41 + 16x².

(b) The derivative of arcsin u is u′/√(1 − u²). So the answer is 2x√(1 − x⁴).

Problem 16 Second derivative of a sum

f′(x) = 5x⁴ − 12x² + 2e^2x. Then f″(x) = 20x³ − 24x + 4e^2x. At 0: 0 − 0 + 4 = 4.

Problem 17 Second derivative of an implicit curve

First, dydx = −x/y. Then d²ydx² = −y − xdydxy² = −y + x²/yy² = −x² + y²y³ = −36y³. At (0, 6): −36/216 = −1/6. The curve is concave down there.

Problem 18 Pressure on a climbing balloon

By the chain rule, dP/dt = P′(h)·h′(t) = −20e^−0.2h·3 = −60e^−0.2h. At t = 2, h = 11, so dP/dt = −60e^−2.2 ≈ −6.65 kPa per hour.

Problem 19 Camera tracking a rocket

The derivative of arctan u is u′/(1 + u²) with u = x/20 and u′ = 1/20. So dθdx = 1/201 + x²/400 = 20400 + x². At x = 15: 20/625 = 4/125 ≈ 0.032 radian per foot.

Problem 20 Horizontal and vertical tangents

Differentiate: 2x − y − xdydx + 2ydydx = 0, so dydx = y − 2x2y − x.

Horizontal: y = 2x. Substitute: x² − 2x² + 4x² = 3x² = 3, so x = ±1 and the points are (1, 2) and (−1, −2).

Vertical: x = 2y. Substitute: 4y² − 2y² + y² = 3y² = 3, so y = ±1 and the points are (2, 1) and (−2, −1).