How do you differentiate #f(x)=(x^2-3x+2)/(x-3)# using the quotient rule?

1 Answer
Steve M
Nov 3, 2016

# f'(x) = ( x^2-6x+7 ) / (x-3)^2 #

Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

# d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #, or less formally, # (u/v)' = (v(du)-u(dv))/v^2 #

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with # f(x)=(x^2-3x+2)/(x-3) # we have;

# f'(x) = {( (x-3)(d/dx(x^2-3x+2)) - (x^2-3x+2)(d/dx(x-3)) )} / (x-3)^2 #
# f'(x) = ( (x-3)(2x-3) - (x^2-3x+2)(1) ) / (x-3)^2 #
# f'(x) = ( 2x^2-9x+9 - x^2+3x-2 ) / (x-3)^2 #
# f'(x) = ( x^2-6x+7 ) / (x-3)^2 #

Related questions
Impact of this question
460 views around the world
You can reuse this answer
Creative Commons License