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#