Question

Find StartFraction dy Over dx EndFraction for y equals StartFraction negative 5 x cubed minus 5 x squared plus 3 Over negative 5 x Superscript 4 Baseline plus 2 EndFraction . Do not simplify.

Answer 1

`(dy)/(dx)=((-5x^4+2)(-15x^2-10x)+(-5x^3-5x^2+3)(-20x^3))/((-5x^4+2)^2)`

Explanation:

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

We are required to find the derivative of y with respect to x,
`(dy)/(dx)` .

We apply the quotient rule:

For a fractional expression,

`(u)/(v),$ (dy)/(dx)=(vu'+uv')/(v^2)`

`u=-5x^3-5x^2+3,$ $u'=-15x^2-10x\\\\v=-5x^4+2, v'=-20x^3`

`(dy)/(dx)=((-5x^4+2)(-15x^2-10x)+(-5x^3-5x^2+3)(-20x^3))/((-5x^4+2)^2)`

Since we are asked not to simplify, we simply leave our answer in the substituted form above.