Question

How do you differentiate `y=(x^3-x^2-3)/(x^5+3)` using the quotient rule?

Answer 1

`(dy)/(dx)=(x^4+15x^2-4x)/(x^2+5)^2`

Explanation:

Quotient rule states if `y=y(x)=(g(x))/(h(x))`

then `(dy)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2`

Here `g(x)=x^3-x^2-3` hence `(dg)/(dx)=3x^2-2x`

and as `h(x)=x^2+5`, `(dh)/(dx)=2x`

Hence `(dy)/(dx)=((3x^2-2x)xx(x^2+5)-2x xx (x^3-x^2-3))/(x^2+5)^2`

= `(3x^4-2x^3+15x^2-10x-2x^4+2x^3+6x)/(x^2+5)^2`

= `(x^4+15x^2-4x)/(x^2+5)^2`