1. Lesson: _____ Section 3.6 Derivatives of Inverse Functions
arcsin =
Range of arcsin and anctan is [− 𝜋 2 , 𝜋 2 ]
Range of arccos is [0,𝜋]
Review:
sin 𝑥 ′ =
cos 𝑥 ′ =
tan 𝑥 ′ =
arccos −1 2 =
arc𝑡𝑎𝑛 3 =
𝝅 𝟔
𝒔𝒊𝒏𝒙
𝟏 𝟐
𝒔𝒊 𝒏 −𝟏 𝒙

2. Explore this on the calculator by graphing:
Derivative of 𝒍𝒏⁡𝒙
Explore this on the calculator by graphing:
y1 = ln x and
y2 = nderiv (y1, x, x)
We know that 𝑒 𝑙𝑛𝑥 =𝑥
then 𝒅 𝒅𝒙 𝑒 𝑙𝑛𝑥 = 𝒅 𝒅𝒙 𝑥
(𝐞 𝐥𝐧𝐱 ) 𝒅 𝒅𝒙 𝐥𝐧 𝒙 =𝟏
𝒅 𝒅𝒙 𝐥𝐧 𝒙 = 𝟏 𝒆 𝒍𝒏𝒙
Note: 𝒙>𝟎 The domain of the derivative can’t exceed the domain of the original function. (If there aren’t any points, there can’t be any slopes!)
𝒅 𝒅𝒙 𝐥𝐧 𝒙 = 𝟏 𝒙
Ex. 𝒅 𝒅𝒙 ln⁡( 𝑥 2 +1)=

3. Derivative of 𝑰𝒏𝒗𝒆𝒓𝒔𝒆 𝑻𝒓𝒊𝒈 𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏𝒔
See bottom of p. 139 for a proof similar to the one we just did for ln⁡(𝑥).
𝒅 𝒅𝒙 𝒂𝒓𝒄𝒕𝒂𝒏 𝒙 = 𝟏 𝟏+ 𝒙 𝟐
𝒅 𝒅𝒙 𝒂𝒓𝒄𝒔𝒊𝒏 𝒙 = 𝟏 𝟏− 𝒙 𝟐
Ex. 𝒅 𝒅𝜽 𝒂𝒓𝒄𝒔𝒊𝒏(𝒕𝒂𝒏𝜽)
𝒅 𝒅𝒙 𝒂𝒓𝒄𝒄𝒐𝒔 𝒙 = −𝟏 𝟏− 𝒙 𝟐

4. Derivative of 𝐚𝐧 𝐈𝐧𝐯𝐞𝐫𝐬𝐞 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝒊𝒏 𝑮𝒆𝒏𝒆𝒓𝒂𝒍
If f and g are inverse functions, then
We know that 𝑓 𝑔 𝑥 =𝑥
then 𝒅 𝒅𝒙 𝑓(𝑔 𝑥 = 𝒅 𝒅𝒙 𝑥
𝑓′(𝑔 𝑥 )∙ 𝑔 ′ (𝑥)=1
Ex. 𝒅 𝒅𝒙 𝒂𝒓𝒄𝒔𝒊𝒏(𝒕𝒂𝒏𝜽)
𝑔 ′ (𝑥)= 1 𝑓 ′ 𝑔 𝑥
(𝑓 −1 )′(𝑥)= 1 𝑓 ′ 𝑓 −1 𝑥
In other words…
“The derivative of the inverse is equal to the reciprocal of the derivative of the original function (evaluated at the point 𝑓 −1 (𝑥) rather than x).”

5. Let’s try to visualize this using the functions below
The derivative of the inverse of a function
(𝑓 −1 )′(𝑥)= 1 𝑓 ′ 𝑓 −1 𝑥
m=12
𝒇 −𝟏 ′ 8 = 1 ( 𝒇 ′ 𝒇 −𝟏 𝟖
(2,8)
m=1/12
𝒇 −𝟏 ′ 8 = 1 ( 𝒇 ′ ( ))
= 𝟏 𝟏𝟐
(8,2) 2
𝒇 −𝟏 (𝒙)= 𝟑 𝒙
Note that the slope of the inverse at 8 is equal to the reciprocal of the slope of the original function at 2.
𝒇(𝒙)=𝒙 𝟑

6. Ex. Given that f and g are differentiable everywhere,
g is the inverse of f, and that 𝑓(3)=4, 𝑓’(3)=6, 𝑓’(4)=7, find 𝑔’(4).
1 𝑓′( )
= 𝟏 𝟔
𝑔 ′ 4 = 3
Rembember, the slopes are reciprocals not at the same x value, but at the corresponding value on the inverse.
The question is, “who is function g associating with 4?” or “Which input for f corresponds to an input of 4 for g?”