How do you differentiate #f(x)=(x+3)^2/(x-1)# using the quotient rule?

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Answer 1 · 2016-07-22T14:13:35

You have to first differentiate the numerator using the chain rule.

Let #y = u^2# and #u = x + 3#. Then #y' = 2u# and #u' = 1. The derivative of the numerator is :

#dy/dx = 2u xx 1 = 2(x + 3) = 2x + 6#

Now, we can use the quotient rule to differentiate the entire function.

#f'(x) = ((2x + 6)(x - 1) - 1(x + 3)^2)/(x - 1)^2#

#f'(x) = (2x^2 + 6x - 2x - 6 - 1(x^2 + 6x + 9))/(x^2 - 2x + 1)#

#f'(x) = (x^2 - 2x -15)/(x^2 - 2x + 1)#

#f'(x) = ((x - 5)(x + 3))/(x - 1)^2#

Hopefully this helps!