What is the derivative of this function y=sin^-1(x/3)?

1 Answer
NickTheTurtle
Apr 6, 2017

1/(sqrt(9-x^2))

Explanation:

I am going to assume that the sin^-1(x) mean arcsin(x) instead of csc(x).

Looking at the differentiation of trigonometric functions, the derivative of arcsin(x) is 1/sqrt(1-x^2).

Proof:
If y=arcsin(x), then x=sin(y). Differentiating both sides with respect to x, we get 1=cos(y)dy/dx. Then, dy/dx=1/cos(y), which is equal to 1/sqrt(1-sin^2(y)). But we said that y=arcsin(x), so dy/dx=1/sqrt(1-x^2).

In order to differentiate arcsin(x/3), we use the chain rule: dy/dx=dy/(du)*(du)/dx, with y=arcsin(x/3) and u=x/3.

Then, d/dx(arcsin(x/3))=d/(d(x/3))(arcsin(x/3))*d/dx(x/3)=1/sqrt(1-(x/3)^2)*1/3=1/(3sqrt(1-1/9x^2))=1/(sqrt(3^2(1-1/9x^2)))=1/(sqrt(9-x^2)).

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
8805 views around the world
You can reuse this answer
Creative Commons License