How do you find the derivative of f(x)=(x^3-3x^2+4)/x^2f(x)=x33x2+4x2?

1 Answer
Nov 28, 2016

The answer is =1-8/x^3=18x3

Explanation:

This is the derivative of a quotient

(u/v)'=(u'v-uv')/v^2

Here,

u=x^3-3x^2+4, =>, u'=3x^2-6x

v=x^2, =>, v'=2x

f'(x)=((3x^2-6x)x^2-(x^3-3x^2+4)2x)/x^4

=(3x^4-6x^3-2x^4+6x^3-8x)/x^4

=(x^4-8x)/x^4

=1-8/x^3

Second possibility,

f(x)=x-3+4x^(-2)

f'(x)=1-4*-2x^(-3)

f'(x)=1-8/x^3