AP Calculus AB · Lesson 3

Inverse functions and inverse trig

Know Know it

If gg is the inverse of ff and f(a)=bf(a) = b, then g(b)=ag(b) = a. Differentiating f(g(x))=xf(g(x)) = x with the chain rule gives f′(g(x)) g′(x)=1f'(g(x))\,g'(x) = 1. So g′(b)=1f′(a)g'(b) = \frac{1}{f'(a)}, as long as f′(a)≠0f'(a) \ne 0.

In words: the slope of the inverse at (b,a)(b, a) is the reciprocal of the slope of ff at the matching point (a,b)(a, b). The hard part is finding aa. Solve f(a)=bf(a) = b, or look for bb in the output column of a table.

The inverse trig derivatives: ddxarcsin⁡x=11−x2\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}, ddxarccos⁡x=−11−x2\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}, and ddxarctan⁡x=11+x2\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}. With an inside function, the chain rule adds a factor: ddxarctan⁡(u)=u′1+u2\frac{d}{dx}\arctan(u) = \frac{u'}{1 + u^2}.

Worked example Easy

What is ddxarctan⁡(2x)\frac{d}{dx}\arctan(2x)?

  1. 11+4x2\frac{1}{1 + 4x^2}
  2. 21+4x2\frac{2}{1 + 4x^2}Answer
  3. 21+2x2\frac{2}{1 + 2x^2}
  4. 21−4x2\frac{2}{\sqrt{1 - 4x^2}}

How to solve it

  1. Inside u=2xu = 2x, so u′=2u' = 2.
  2. ddxarctan⁡u=u′1+u2=21+(2x)2\frac{d}{dx}\arctan u = \frac{u'}{1 + u^2} = \frac{2}{1 + (2x)^2}.
  3. Simplify: 21+4x2\frac{2}{1 + 4x^2}.

Why each choice is right or wrong

  • A. Incorrect. 11+4x2\frac{1}{1 + 4x^2} forgets the chain rule factor 2 from the inside function 2x2x.
  • B. Correct. ddxarctan⁡u=u′1+u2\frac{d}{dx}\arctan u = \frac{u'}{1 + u^2} with u=2xu = 2x, so it is 21+(2x)2=21+4x2\frac{2}{1 + (2x)^2} = \frac{2}{1 + 4x^2}.
  • C. Incorrect. 21+2x2\frac{2}{1 + 2x^2} squares only the xx; the whole inside 2x2x is squared, giving 4x24x^2.
  • D. Incorrect. 21−4x2\frac{2}{\sqrt{1 - 4x^2}} is the derivative of arcsin⁡(2x)\arcsin(2x), not of arctan⁡(2x)\arctan(2x).

Worked example Medium

Let f(x)=x3+2x−1f(x) = x^3 + 2x - 1, and let gg be the inverse function of ff.

What is g′(2)g'(2)?

  1. 114\frac{1}{14}
  2. 15\frac{1}{5}Answer
  3. 55
  4. 111\frac{1}{11}

How to solve it

  1. Find aa with f(a)=2f(a) = 2: try a=1a = 1, and 1+2−1=21 + 2 - 1 = 2. So g(2)=1g(2) = 1.
  2. f′(x)=3x2+2f'(x) = 3x^2 + 2, so f′(1)=5f'(1) = 5.
  3. g′(2)=1f′(1)=15g'(2) = \frac{1}{f'(1)} = \frac{1}{5}.

Why each choice is right or wrong

  • A. Incorrect. 114\frac{1}{14} is 1f′(2)\frac{1}{f'(2)}: it uses the input 2 instead of the input aa with f(a)=2f(a) = 2.
  • B. Correct. f(1)=1+2−1=2f(1) = 1 + 2 - 1 = 2, so g(2)=1g(2) = 1. Since f′(x)=3x2+2f'(x) = 3x^2 + 2, f′(1)=5f'(1) = 5, and g′(2)=1f′(1)=15g'(2) = \frac{1}{f'(1)} = \frac{1}{5}.
  • C. Incorrect. 5 is f′(1)f'(1); the derivative of the inverse is its reciprocal.
  • D. Incorrect. 111\frac{1}{11} is 1f(2)\frac{1}{f(2)}, the reciprocal of a function value, not of a slope.

Explain Explain it

Task verbs: explain, calculate

With a table, find the row where f(x)f(x) equals the input you were given for gg. That row's xx is aa, and g′(b)=1f′(a)g'(b) = \frac{1}{f'(a)}. Do not use the row where xx equals bb.

On the graphs, ff and its inverse are reflections over y=xy = x. The point (a,b)(a, b) on ff matches (b,a)(b, a) on gg. Reflecting a line of slope mm over y=xy = x gives a line of slope 1m\frac{1}{m}, which is why the slopes are reciprocals.

Trap Use the matching point, not the same number

(f−1)′(5)(f^{-1})'(5) is 1f′(a)\frac{1}{f'(a)} where f(a)=5f(a) = 5. It is not 1f′(5)\frac{1}{f'(5)}. Students who plug 5 into f′f' get a wrong answer that is usually one of the choices.

Worked example Hard

The function ff is differentiable and one-to-one, and gg is its inverse. Selected values of ff and f′f' are shown in the table.

Values of ff and f′f'

xx1234
f(x)f(x)24711
f′(x)f'(x)3526

Let h(x)=x g(x)h(x) = x\,g(x). What is h′(4)h'(4)?

  1. 145\frac{14}{5}Answer
  2. 83\frac{8}{3}
  3. 2611\frac{26}{11}
  4. 2222

How to solve it

  1. Find the row where f(x)=4f(x) = 4: it is x=2x = 2, so g(4)=2g(4) = 2.
  2. g′(4)=1f′(2)=15g'(4) = \frac{1}{f'(2)} = \frac{1}{5}.
  3. Product rule: h′(4)=1⋅g(4)+4⋅g′(4)=2+45=145h'(4) = 1 \cdot g(4) + 4 \cdot g'(4) = 2 + \frac{4}{5} = \frac{14}{5}.

Why each choice is right or wrong

  • A. Correct. f(2)=4f(2) = 4, so g(4)=2g(4) = 2 and g′(4)=1f′(2)=15g'(4) = \frac{1}{f'(2)} = \frac{1}{5}. By the product rule, h′(4)=g(4)+4g′(4)=2+45=145h'(4) = g(4) + 4g'(4) = 2 + \frac{4}{5} = \frac{14}{5}.
  • B. Incorrect. 83\frac{8}{3} uses g′(4)=1f′(4)=16g'(4) = \frac{1}{f'(4)} = \frac{1}{6}; the slope must come from the row where f(x)=4f(x) = 4, which is x=2x = 2.
  • C. Incorrect. 2611\frac{26}{11} uses g′(4)=1f(4)=111g'(4) = \frac{1}{f(4)} = \frac{1}{11}, the reciprocal of a function value rather than of the slope f′(2)=5f'(2) = 5.
  • D. Incorrect. 22 uses g′(4)=f′(2)=5g'(4) = f'(2) = 5; the derivative of the inverse is the reciprocal, 15\frac{1}{5}.

Explain Your turn

The function ff is differentiable and one-to-one, with f(6)=2f(6) = 2 and f′(6)=4f'(6) = 4. Let gg be the inverse of ff. Find g′(2)g'(2), and explain why you use f′(6)f'(6) rather than f′(2)f'(2).

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A full answer does this

  • Finds g′(2)=14g'(2) = \frac{1}{4}
  • Explains that g(2)=6g(2) = 6, so the matching point on ff is at x=6x = 6

Model answer

Since f(6)=2f(6) = 2, the inverse gives g(2)=6g(2) = 6. The point (2,6)(2, 6) on gg matches the point (6,2)(6, 2) on ff, and the slopes at matching points are reciprocals. So g′(2)=1f′(6)=14g'(2) = \frac{1}{f'(6)} = \frac{1}{4}. Using f′(2)f'(2) would take the slope of ff at a point that does not match (2,6)(2, 6).

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