Question

How do you differentiate `f(x)=sqrt(((3x)/(2x-3))` using the chain rule?

Answer 1

`f'(x)=-9/2*((3*x)/(2*x-3))^(-1/2)`

Explanation:

Writing

`f(x)=((3*x)/(2x-3))^(1/2)`

so we get

`f'(x)=1/2*((3*x)/(2x-3))^(-1/2)*(3*(2x-3)-3x*2)/(2x-3)^2`

so

`f'(x)=-9/2*((3x)/(2x-3))^(-1/2)`