Question

find the derivative of x^2 cosec^-1 (1/lnx)​

Answer 1

Step-by-step explanation:

Introduction to Inverse Trigonometric Functions

In the previous topic, we have learned the derivatives of six basic trigonometric functions:

sinx,cosx,tanx,cotx,secx,cscx.

In this section, we are going to look at the derivatives of the inverse trigonometric functions, which are respectively denoted as

arcsinx,arccosx,arctanx,arccot x,arcsec x,arccsc x.

The inverse functions exist when appropriate restrictions are placed on the domain of the original functions.

For example, the domain for arcsinx is from −1 to 1. The range, or output for arcsinx is all angles from −π2 to π2 radians.

The domains of the other trigonometric functions are restricted appropriately, so that they become one-to-one functions and their inverse can be determined.

Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions can be obtained using the inverse function theorem. For example, the sine function x=φ(y) =siny is the inverse function for y=f(x) =arcsinx. Then the derivative of y=arcsinx is given by

by

(arcsinx)′=f′(x)=1φ′(y)=1(siny)′=1cosy=1√1−sin2y=1√1−sin2(arcsinx)=1√1−x2(−1<x<1).(arcsin⁡x)′=f′(x)=1φ′(y)=1(sin⁡y)′=1cos⁡y=11−sin2y=11−sin2(arcsin⁡x)=11−x2(−1<x<1).

Using this technique, we can find the derivatives of the other inverse trigonometric functions:

functions:

(arccosx)′=1(cosy)′=1(−siny)=−1√1−cos2y=−1√1−cos2(arccosx)=−1√1−x2(−1<x<1),(arccos⁡x)′=1(cos⁡y)′=1(−sin⁡y)=−11−cos2y=−11−cos2(arccos⁡x)=−11−x2(−1<x<1),

(arctanx)′=1(tany)′=11cos2y=11+tan2y=11+tan2(arctanx)=11+x2,(arctan⁡x)′=1(tan⁡y)′=11cos2y=11+tan2y=11+tan2(arctan⁡x)=11+x2,

(arccot x)′=1(coty)′=1(−1sin2y)

Table of Derivatives of Inverse Trigonometric Functions

The derivatives of 66 inverse trigonometric functions considered above are consolidated in the following table:



In the examples below, find the derivative of the given function.

Example 1

y=arctan1xy=arctan⁡1x

Example 2

y=arcsin(x−1)y=arcsin⁡(x−1)

Example 3

y=arccotx2y=arccotx2

Example 4

y=1aarctanxay=1aarctan⁡xa

Example 5

y=arctanx+1x−1(x≠1)y=arctan⁡x+1x−1(x≠1)

Example 6

y=arccot1x2y=arccot1x2

Example 7

y=arctan(x−√1+x2)y=arctan⁡(x−1+x2)

Example 8

Calculate the derivative of the function y=arccosxarctanxy=arccos⁡xarctan⁡x at x=0.x=0.

Example 9

Using the chain rule, derive the formula for the derivative of the inverse sine function.

Example 10

y=arcsin√1−x2y=arcsin⁡1−x2

Example 11

Calculate the derivative of y=arctan√exy=arctan⁡ex at x=0.x=0.

Example 12

Compute the derivative of the function y=arcsin1√xy=arcsin1x at x=2.x=2.

Example 13

y=2√3arccot2x+1√3y=23arccot2x+13

Example 14

Show that ddx

Example 1.

y=arctan1xy=arctan⁡1x

Solution.

By the chain rule,

y′=(arctan1x)′=11+(1x)2⋅(1x)′=11+1x2⋅(−1x2)=−x2(x2+1)x2=−11+x2.y′=(arctan⁡1x)′=11+(1x)2⋅(1x)′=11+1x2⋅(−1x2)=−x2(x2+1)x2=−11+x2.

Example 2.

y=arcsin(x−1)y=arcsin⁡(x−1)

Solution.

y′=(arcsin(x−1))′=1√1−(x−1)2=1√1−(x2−2x+1)=1√1−x2+2x−1=1√2x−x2