Lesson: _____ Section 3.6 Derivatives of Inverse Functions arcsin 3 2 = 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

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)=

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

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).”

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

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?”