AP Calculus AB · Free-response skill

Derivatives from a table

The move. From a table of values of two functions and their derivatives, find derivatives of a composition, a product and an inverse at a point, and a slope on a curve defined implicitly.

How to do it

  1. Write the rule with function names first, then substitute. A reader can follow a line such as "the outer derivative at g(2)g(2), times g′(2)g'(2)" even if a number slips.
  2. For a composition f(g(x))f(g(x)) at aa, find g(a)g(a) in the table first, then read f′f' at that input. Look up the inside before the outside.
  3. For an inverse at bb, find the row where the function's output is bb. If f(a)=bf(a) = b, then (f−1)′(b)=1f′(a)(f^{-1})'(b) = \frac{1}{f'(a)}.
  4. For an equation that mixes xx and yy, differentiate every term, attach dydx\frac{dy}{dx} to each yy term, and solve for dydx\frac{dy}{dx} before you substitute the point.
  5. Every step here can be done by hand, so practice it without a calculator.

Practice prompt Try it

The functions ff and gg are differentiable for all real numbers, and gg is one-to-one. Selected values of ff, gg and their derivatives are shown in the table.

Values of ff, gg and their derivatives

xxf(x)f(x)f′(x)f'(x)g(x)g(x)g′(x)g'(x)
13-225
2-1443
45672

Use the table to answer each part. Show the work that leads to each answer.

  1. Let h(x)=f(g(x))h(x) = f(g(x)). Find h′(1)h'(1).
  2. Let k(x)=[f(x)]2g(x)k(x) = [f(x)]^2 g(x). Find k′(2)k'(2).
  3. Let g−1g^{-1} be the inverse function of gg. Find (g−1)′(4)(g^{-1})'(4).
  4. A curve is given by y2+xy=g(x)y^2 + xy = g(x). The point (1,1)(1, 1) is on the curve. Find dydx\frac{dy}{dx} at the point (1,1)(1, 1).

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Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 7 points

Model answer, annotated

(a) By the chain rule, h′(1)=f′(g(1))⋅g′(1)h'(1) = f'(g(1)) \cdot g'(1). From the table g(1)=2g(1) = 2, so h′(1)=f′(2)⋅g′(1)=4⋅5=20h'(1) = f'(2) \cdot g'(1) = 4 \cdot 5 = 20.

  • Point 1 States the chain rule with the inside g(1) read first.
  • Point 2 Uses f'(2) = 4 and g'(1) = 5 to get 20.

(b) By the product and chain rules, k′(x)=2f(x)f′(x) g(x)+[f(x)]2g′(x)k'(x) = 2f(x)f'(x)\,g(x) + [f(x)]^2 g'(x). At x=2x = 2: k′(2)=2(−1)(4)(4)+(−1)2(3)=−32+3=−29k'(2) = 2(-1)(4)(4) + (-1)^2(3) = -32 + 3 = -29.

  • Point 3 Writes the product rule with the chain rule on the square.
  • Point 4 Substitutes the x = 2 row of the table to get -29.

(c) The table shows g(2)=4g(2) = 4, so g−1(4)=2g^{-1}(4) = 2. Then (g−1)′(4)=1g′(2)=13(g^{-1})'(4) = \frac{1}{g'(2)} = \frac{1}{3}.

  • Point 5 Finds the matching input 2 from g(2) = 4 and takes the reciprocal of g'(2).

(d) Differentiating both sides with respect to xx gives 2ydydx+y+xdydx=g′(x)2y\frac{dy}{dx} + y + x\frac{dy}{dx} = g'(x). At (1,1)(1, 1), with g′(1)=5g'(1) = 5: 2dydx+1+dydx=52\frac{dy}{dx} + 1 + \frac{dy}{dx} = 5, so 3dydx=43\frac{dy}{dx} = 4 and dydx=43\frac{dy}{dx} = \frac{4}{3}.

  • Point 6 Differentiates implicitly, with dy/dx on the y-squared term and the product rule on xy.
  • Point 7 Substitutes the point and g'(1) = 5 and solves to get 4/3.