How do you find the derivative of f(x)=x^3/(x^3-5x+10)?

1 Answer
Mar 25, 2018

=-10(x^2(x-3))/(x^3-5x+10)^2

Explanation:

Use the quotient rule:

Let u=f(x) and v=g(x)

Then d/dx(u/v)=(u'v-uv')/v^2

For this question:

u=x^3
u'=3x^2

v=x^3-5x+10
v'=3x^2-5

d/dx(u/v)=(3x^2(x^3-5x+10)-x^3(3x^2-5))/(x^3-5x+10)^2

=(cancel(3x^5)-15x^3+30x^2cancel(-3x^5)+5x^3)/(x^3-5x+10)^2

=(-10x^3+30x^2)/(x^3-5x+10)^2

=-10(x^2(x-3))/(x^3-5x+10)^2