You are here:

Calculus/calculus proof

Advertisement

Expert: - 4/7/2010


Question
I need a step by step proof of
(sin^-1(x)=((1-x^2)^(-1/2)or
(sin^-1(x)= 1/(1-x^2)^(1/2)

Answer
I assume you want the derivative of sin^-1(x)

We know that sin(sin^-1(x)) = x

Take the derivative of both sides , and use the chain rule on the left side

cos(sin^-1(x)) (sin^-1(x))'  = 1

so

(sin^-1(x))'  = 1 / cos(sin^-1(x))


We know that sine squared plus cosine squared is always 1 ,so

[sin(sin^-1(x))]^2  + [cos(sin^-1(x))]^2  = 1

x^2 + [cos(sin^-1(x))]^2  = 1

[cos(sin^-1(x))]^2 = 1 - x^2

cos(sin^-1(x)) = (1 - x^2)^1/2

substitute this into (sin^-1(x))'  = 1 / cos(sin^-1(x)) and get

(sin^-1(x))'  = 1 / (1 - x^2)^1/2

Calculus

All Answers

Answers by Expert:

Ask Experts

Volunteer

Socrates

Expertise

I can answer questions from the standard four semester Calculus sequence. I am not prepared for questions on Tensor Calculus. Everything else is welcome. Derivatives, partial derivatives, ordinary differential equations, single and multiple integrals, change of variable, vector integration (Green`s Theorem, Stokes, and Gauss) and applications.

Experience

Ph.D. in Mathematics and many years teaching Calculus at state universities.

Education/Credentials
B.S. , M.S. , Ph.D.

©2012 About.com, a part of The New York Times Company. All rights reserved.