Question

How do you differentiate `y = (5x^2) / (4x-1)`?

Answer 1

`dy/dx = (10x(2x-1))/ (4x-1)^2`

Just use the quotient rule and simplify.

Explanation:

`dy/dx = ((4x-1)* d/dx [ 5x^2 ] - 5x^2 * d/dx [ 4x-1 ] )/ (4x-1)^2`

`dy/dx = ((4x-1)* 10x - 5x^2 * 4 )/ (16x^2+1-8x)`

`dy/dx = ((40x^2-10x) - 20x^2 )/ (16x^2+1-8x)`

`dy/dx = (10x(2x-1))/ (4x-1)^2`