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

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Answer 1 · 2016-11-28T08:51:36

The answer is $=1-8/x^3$

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,

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

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

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

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