Question

What is the derivative of `y = Arcsin ((3x)/4)`?

Answer 1

Use the Chain rule and the definition of the derivative of the arcsin of a function.

Explanation:

First, understand that `y=sin^-1(3x/4)` is way of saying `y=f(g(x))`

Second, let `f(m) = sin^-1(m)` (i.e., the "outside" part) and `m=g(x)=3x/4` (i.e., the "inside" part).

Third, use the Chain Rule which states `dy/dx=dy/(dm)*(dm)/(dx)=f'(m)g'(x)`

Fourth, find `(dy)/(dm)`

`(dy)/(dm)= f'(m) = d/(dm)(sin^-1(m))=1/sqrt(1-x^2)`

Fifth, find `(dm)/(dx)`

`(dm)/(dx)=g'(x)=d/(dx)(3x/4)=3/4`

Finally, multiply your results:

`(dy)/(dx)=(dy)/(dm)(dm)/(dx)=1/sqrt(1-x^2)*3/4=3/(4sqrt(1-x^2))`.

Done, and done!