Question

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

Answer 1

`dy/dx=(3(x^2-3x))/((2x-3)^2sqrt((3x^2)/(2x-3))`

Explanation:

Differentiate the outer function and multiply by the derivative of the inside like so:

`dy/dx=1/2((3x^2)/(2x-3))^(-1/2)*d/dx[(3x^2)/(2x-3)]`

Evaluating the derivative of the inside function we get:

`dy/dx=1/2((3x^2)/(2x-3))^(-1/2)((2x-3)(6x)-(3x^2)2)/(2x-3)^2`

Simplify:

`dy/dx=(6(x^2-3x))/(2(2x-3)^2sqrt((3x^2)/(2x-3))`

Cancel 2's:

`dy/dx=(3(x^2-3x))/((2x-3)^2sqrt((3x^2)/(2x-3))`

Technically, it would be proper to rationalize the denominator, but you can do that on your own.