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

1 Answer
Oct 18, 2017

Answer:

#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.