1
Derivatives of inverse function
𝑓(𝑔(𝑥)) = 𝑥
.
𝑓′(𝑔(𝑥))𝑔′(𝑥) = 1
𝑔′ (𝑥) =
1
𝑓′(𝑔(𝑥))
The beauty of this formula is that we don’t need to actually determine 𝑔(𝑥) to find the value of
the derivative at a point. We simply use the reflection property of inverse function:
Derivative of the inverse function at a point is the reciprocal of the derivative of the
function at the corresponding point.
Slope of the line tangent to 𝒇−𝟏 at 𝒙 = 𝒃 is the reciprocal of the slope of 𝒇 at 𝒙 = 𝒂.
1. Find tangent line at point (4, 2) of the graph of f -1 if f(x) = x3 + 2x – 8
2. Find the equation of the tangent line to the inverse at the given point.
a. f(x) = x3 + 7x +2
b. f(x) = x5 + 3x3 + 7x +2
@ (10, 1)
@ (13, 1)
2
c. f(x) = e-2x – 9x3 + 4
d. f(x) = x7 + 2x + 9
e. f(x) = x5/3 𝑒 𝑥
𝑓.
𝑓(𝑥) =
2
−𝑒 −3𝑥
𝑥2 + 1
g. f(x) = 7x + sin (2x)
@ (5, 0)
@ (12, 1)
@ (e, 1)
@ (−1, 0)
@ (0,0)
3
h. f(x) = x3 + 8x + cos (3x)
i. f(x) = 10x + (arc tanx)2
j. f(x) = 7x3 + (ln x)3
@ (1,0)
@ (0,0)
@ (7, 1)
3. A function 𝑓 and its derivative take on the values shown in the table. If 𝑔 is the inverse of 𝑓,
find 𝑔′ (6).
x
2
6
f(x)
6
8
f’(x)
1/3
3/2
4
4. Let y = f(x) = x3 + x + x – 2, and let g be the inverse function. Evaluate g’(0).
SOLUTIONS
1.
𝑓(𝑦(𝑥)) = 𝑥
𝑓′(𝑦(𝑥))𝑦′(𝑥) = 1
𝑦 ′ (4) =
f’ = 3x2 + 2
2. a y – 1 = 1/10 (x – 10)
2. b
y – 1 = 1/21 (x – 13)
2. c
y = - ½ (x – 5)
2. d
y – 1 = 1/9 (x – 12)
2. e
y–1=
1
2. f
y = 1/3 (x + 1)
2. g y = 1/9 x
2. h
y = 1/8 (x-1)
2. i y = 1/10 x
2. j y – 1 = 1/21 (x – 7 )
(2𝑒 + 5/3 𝑒)
3. 𝑓(𝑔(𝑥)) = 𝑥
(x – e)
𝑓′(𝑔(𝑥))𝑔′(𝑥) = 1
f-1(4) =2
𝑓′(𝑦(4))𝑦′(4) = 1
1
𝑓′(2)
= 1/14
𝑓′(𝑔(6))𝑔′(6) = 1
𝑓′(2)𝑔′(6) = 1
𝑔′ (6) = 3
4. f’(x) = 3x2 + 1
𝑔′ (𝑦) =
1
3𝑥 2 +1
g’(0) = 1/(3(1)2 + 1) = 1/4
x=? when x3 + x – 2 = 0
⇒ x=1