Question

Y equals the quotient of the quantity x squared plus 4 times x and the quantity x cubed minus 5.

Answer 1

Given:

Consider the completer question is "Find the derivative
`(dy)/(dx)` for
`y=(x^2-4x)/(x^3-5)`."

To find:

The derivative
`(dy)/(dx)`.

Solution:

Chain rule:
`(d)/(dx)f(g(x))=f'(g(x))\cdot g'(x)`

Quotient rule:
`(d)/(dx)(f(x))/(g(x))=(g(x)f'(x)-f(x)g'(x))/([g(x)]^2)`

We have,

`y=(x^2-4x)/(x^3-5)`

Differentiate with respect to x.

`(dy)/(dx)=(d)/(dx)\left((x^2-4x)/(x^3-5)\right)`

Using chain rule and quotient rule, we get

`(dy)/(dx)=((x^3-5)(d)/(dx)(x^2-4x)-(x^2-4x)(d)/(dx)(x^3-5))/((x^3-5)^2)`

`(dy)/(dx)=((x^3-5)(2x-4)-(x^2-4x)(3x^2))/((x^3-5)^2)`

`(dy)/(dx)=(2x^4-4x^3-10x+20-3x^4+12x^3)/((x^3-5)^2)`

`(dy)/(dx)=(-x^4+8x^3-10x+20)/((x^3-5)^2)`

Therefore, the required answer is
`(dy)/(dx)=(-x^4+8x^3-10x+20)/((x^3-5)^2)`.