Unit 3 · Topic 3.3 Beta

Derivatives of Inverse Functions

The graph of an inverse is a reflection in y = x , which swaps rise and run, so slopes become reciprocals at matching points: (f⁻¹)′(a) = 1/(f′(f⁻¹(a))) .

Practice 1: Implementing Mathematical ProcessesPractice 2: Connecting Representations

Question set for this topic

Part 1 · Hook

Why this matters

If you know how fast a function climbs, you also know how fast its inverse climbs: undo the function and the steepness flips. A thermometer that converts a voltage to a temperature, or a calculator's square root key, is an inverse function, and this topic finds its slope without a formula for it.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If f(2)=7f(2) = 7 and gg is the inverse of ff, what is g(7)g(7)?

  1. 22
  2. 77
  3. 17\frac{1}{7}
  4. 12\frac{1}{2}
Show the answer

An inverse swaps inputs and outputs: f(2)=7f(2) = 7 means g(7)=2g(7) = 2.

  • Correct: 22:
  • 77:
  • 17\frac{1}{7}:
  • 12\frac{1}{2}:

2. By the chain rule, ddxf(g(x))\frac{d}{dx}f(g(x)) is

  1. f′(g(x))⋅g′(x)f'(g(x)) \cdot g'(x)
  2. f′(x)⋅g′(x)f'(x) \cdot g'(x)
  3. f′(g′(x))f'(g'(x))
  4. f(g′(x))f(g'(x))
Show the answer

The outside derivative at the inside function, times the inside derivative.

  • Correct: f′(g(x))⋅g′(x)f'(g(x)) \cdot g'(x):
  • f′(x)⋅g′(x)f'(x) \cdot g'(x):
  • f′(g′(x))f'(g'(x)):
  • f(g′(x))f(g'(x)):

Part 4 · See it

See it first

The graph of f(x) = x³ + x (crimson) and its inverse g (blue), mirror images in the dashed line y = x. The tangent line to f at (1, 2) has slope 4; the tangent line to g at the reflected point (2, 1) has slope 1/4.
Reflecting f(x)=x3+xf(x) = x^3 + x in the line y=xy = x gives its inverse gg. The tangent line at (1,2)(1, 2) has slope 4; its mirror image at (2,1)(2, 1) has slope 14\frac{1}{4}. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. The graph of g=f−1g = f^{-1} is the graph of ff reflected in y=xy = xthe point (a,b)(a, b) on ff becomes (b,a)(b, a) on gg
  2. Reflection in y=xy = x swaps rise and runa tangent line of slope mm becomes one of slope 1m\frac{1}{m}
  3. So the slope of gg at bb is the reciprocal of the slope of ff at a=g(b)a = g(b)g′(b)=1f′(g(b))g'(b) = \frac{1}{f'(g(b))}
  4. Where f′(a)=0f'(a) = 0the tangent line of gg at bb is vertical and g′(b)g'(b) does not exist

Part 6 · Key ideas

Key ideas

  • The rule: g′(x)=1f′(g(x))g'(x) = \frac{1}{f'(g(x))} when gg is the inverse of ff and f′(g(x))≠0f'(g(x)) \ne 0.
  • Two steps: first find g(x)g(x), the input of ff that gives xx; then take the reciprocal of f′f' there.
  • From a table, find the given number in the f(x)f(x) row, read its xx, then read f′f' in that column.
  • It comes from the chain rule: differentiate f(g(x))=xf(g(x)) = x to get f′(g(x))⋅g′(x)=1f'(g(x)) \cdot g'(x) = 1.

Part 7 · Misconception

A common mistake

The wrong idea: If f(2)=4f(2) = 4 and g=f−1g = f^{-1}, then g′(4)=1f′(4)g'(4) = \frac{1}{f'(4)}.

What actually happens: The input of f′f' must be g(4)=2g(4) = 2, the xx where ff gives 4: g′(4)=1f′(2)g'(4) = \frac{1}{f'(2)}. Reading f′f' at 4 uses the wrong point of the graph.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Values of ff and f′f'

Selected values of a differentiable, one-to-one function f and its derivative; g is the inverse function of f
xx0011223344
f(x)f(x)13479
f′(x)f'(x)32243

1. The function gg is the inverse of ff. Find g′(4)g'(4).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

g(4)=2g(4) = 2 because f(2)=4f(2) = 4. Then g′(4)g'(4)=1f′(g(4))= \frac{1}{f'(g(4))}=1f′(2)= \frac{1}{f'(2)}=12= \frac{1}{2}.

  • Answer: 12\frac{1}{2} ≈ 0.500

2. What is g′(7)g'(7)?

  1. 14\frac{1}{4}
  2. 44
  3. 17\frac{1}{7}
  4. 13\frac{1}{3}
Show the answer

Find g(7)g(7) from the table: f(3)=7f(3) = 7, so g(7)=3g(7) = 3. Then g′(7)=1f′(3)=14g'(7) = \frac{1}{f'(3)} = \frac{1}{4}.

  • Correct: 14\frac{1}{4}: g(7)=3g(7) = 3 and f′(3)=4f'(3) = 4, so g′(7)=14g'(7) = \frac{1}{4}.
  • 44: This is f′(3)f'(3); the slope of the inverse is its reciprocal.
  • 17\frac{1}{7}: This is 17\frac{1}{7}, the reciprocal of the input. The rule uses f′f' at g(7)g(7).
  • 13\frac{1}{3}: This is 1g(7)\frac{1}{g(7)}. The rule takes the reciprocal of a slope, f′(g(7))f'(g(7)), not of a value.

3. Which is an equation of the line tangent to the graph of gg at x=3x = 3?

  1. y−1=2(x−3)y - 1 = 2(x - 3)
  2. y−1=12(x−3)y - 1 = \frac{1}{2}(x - 3)
  3. y−3=12(x−1)y - 3 = \frac{1}{2}(x - 1)
  4. y−1=14(x−3)y - 1 = \frac{1}{4}(x - 3)
Show the answer

f(1)=3f(1) = 3 gives g(3)=1g(3) = 1, and g′(3)=1f′(1)=12g'(3) = \frac{1}{f'(1)} = \frac{1}{2}. So y−1=12(x−3)y - 1 = \frac{1}{2}(x - 3).

  • y−1=2(x−3)y - 1 = 2(x - 3): The slope 2 is f′(1)f'(1); the inverse needs its reciprocal.
  • Correct: y−1=12(x−3)y - 1 = \frac{1}{2}(x - 3): The point is (3,g(3))=(3,1)(3, g(3)) = (3, 1) and the slope is 1f′(1)=12\frac{1}{f'(1)} = \frac{1}{2}.
  • y−3=12(x−1)y - 3 = \frac{1}{2}(x - 1): This is a line through (1,3)(1, 3), a point of ff. On gg the coordinates swap: (3,1)(3, 1).
  • y−1=14(x−3)y - 1 = \frac{1}{4}(x - 3): This uses f′(3)=4f'(3) = 4. Read f′f' at g(3)=1g(3) = 1, not at 3.

4. Let h(x)=[g(x)]2h(x) = [g(x)]^2. Find h′(4)h'(4).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

By the chain rule, h′(4)h'(4)=2g(4)⋅g′(4)= 2g(4) \cdot g'(4)=2(2)⋅12= 2(2) \cdot \frac{1}{2}=2= 2, using g(4)=2g(4) = 2 and g′(4)=1f′(2)=12g'(4) = \frac{1}{f'(2)} = \frac{1}{2}.

  • Answer: 2

5. Let f(x)=x5+2x+3f(x) = x^5 + 2x + 3. Find (f−1)′(3)\left(f^{-1}\right)'(3).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

f(0)=3f(0) = 3, so f−1(3)=0f^{-1}(3) = 0. f′(x)=5x4+2f'(x) = 5x^4 + 2, so f′(0)=2f'(0) = 2 and (f−1)′(3)=12\left(f^{-1}\right)'(3) = \frac{1}{2}.

  • Answer: 12\frac{1}{2} ≈ 0.500

6. A one-to-one differentiable function ff has f(2)=5f(2) = 5, f′(2)=−3f'(2) = -3, f(5)=1f(5) = 1 and f′(5)=8f'(5) = 8. What is (f−1)′(5)\left(f^{-1}\right)'(5)?

  1. 18\frac{1}{8}
  2. −13-\frac{1}{3}
  3. −3-3
  4. 12\frac{1}{2}
Show the answer

f(2)=5f(2) = 5 means f−1(5)=2f^{-1}(5) = 2. Then (f−1)′(5)\left(f^{-1}\right)'(5)=1f′(2)= \frac{1}{f'(2)}=−13= -\frac{1}{3}.

  • 18\frac{1}{8}: This is 1f′(5)\frac{1}{f'(5)}. The input 5 is an output of ff; the matching input is 2.
  • Correct: −13-\frac{1}{3}: f−1(5)=2f^{-1}(5) = 2, so the slope is 1f′(2)=−13\frac{1}{f'(2)} = -\frac{1}{3}.
  • −3-3: This is f′(2)f'(2); the inverse's slope is the reciprocal.
  • 12\frac{1}{2}: This is 1f−1(5)\frac{1}{f^{-1}(5)}, the reciprocal of a value instead of a slope.

Part 9 · Summary

Summary

The graph of an inverse is a reflection in y=xy = x, which swaps rise and run, so slopes become reciprocals at matching points: (f−1)′(a)\left(f^{-1}\right)'(a)=1f′(f−1(a))= \frac{1}{f'\left(f^{-1}(a)\right)}. Find f−1(a)f^{-1}(a) first, then take the reciprocal of f′f' at that input.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections