Question

Find the differential of y in the function: y=(x^2-4)/(x-3) ?

Answer 1

`dy/dx=(x^2-6x+4)/(x-3)^2`

Explanation:

We want to find the derivative of

`y=(x^2-4)/(x-3)`

Use the quotient rule, if `y=f/g`

then `dy/dx=(f'*g-f*g')/g^2`, thus

• `f=x^2-4=>f'=2x`
• `g=x-3=>g'=1`

By the quotient rule

`dy/dx=(2x(x-3)-(x^2-4))/(x-3)^2`

`=(2x^2-6x-x^2+4)/(x-3)^2`

`=(x^2-6x+4)/(x-3)^2`