How do you differentiate #f(x) = (x^3-3x^2+4)/x^2#?

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Answer 1 · 2015-04-23T17:53:00

Well, you could use the quotient rule, if you know it, but the easier way (in my opinion) is to think first, differentiate second:

#f(x) = (x^3-3x^2+4)/x^2 = x^3/x^2-(3x^2)/x^2+4/x^2#, so

#f(x) = x-3+4/x^2 #.

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

Therefore: #f'(x) = 1-0-8x^(-3)#

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

you may get a single denominator if you like.