Answer:
[tex]\dfrac{dy}{dx}=\dfrac{(-5x^4+2)(-15x^2-10x)+(-5x^3-5x^2+3)(-20x^3)}{(-5x^4+2)^2}[/tex]
Step-by-step explanation:
Given: [tex]y=\dfrac{-5x^3-5x^2+3}{-5x^4+2}[/tex]
We are required to find the derivative of y with respect to x, [tex]\dfrac{dy}{dx}[/tex] .
We apply the quotient rule:
For a fractional expression,
[tex]\dfrac{u}{v},$ \dfrac{dy}{dx}=\dfrac{vu'+uv'}{v^2}[/tex]
[tex]u=-5x^3-5x^2+3,$ $u'=-15x^2-10x\\\\v=-5x^4+2, v'=-20x^3[/tex]
[tex]\dfrac{dy}{dx}=\dfrac{(-5x^4+2)(-15x^2-10x)+(-5x^3-5x^2+3)(-20x^3)}{(-5x^4+2)^2}[/tex]
Since we are asked not to simplify, we simply leave our answer in the substituted form above.
