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

2 Answers
Monzur R.
Mar 5, 2018

#f'(x)=-(x-9)/(x+3)^3#

Explanation:

We use the quotient rule to find #f'(x)#:

#d/dx u/v=(vu'-uv')/v^2#

So

#d/dx(x-3)/(x+3)^2=((x+3)^2-2(x+3)(x-3))/(x+3)^4=(x+3-2x+6)/(x+3)^3=-(x-9)/(x+3)^3#

Shiva Prakash M V
Mar 5, 2018

#d/dx(f(x))=-(x-9)/(x+3)^3#

Explanation:

#f(x)=(x-3)/(x+3)^2#
let
#y=f(x)#
#y=(x-3)/(x+3)^2#
Let
#u=x-3#
Then
#du/dx=1#
Let
#v=(x+3)^2#
#(dv)/dx=2(x+3)#
and
#v^2=(x+3)^4#
We have by quotient rule in diferentiation

#d/dx(u/v)=(v(du)/dx-u(dv)/dx)/v^2#
Substituting
#d/dx((x-3)/((x+3)^2))=((x+3)^2(1)-(x-3)(2(x+3)))/(x+3)^4#
Simplifying
#d/dx(y)=((x+3)(x+3-2(x-3)))/(x+3)^4#

#d/dx(f(x))=((x+3)(x+3-2x+6))/(x+3)^4#

#=(-x+9)/(x+3)^3#
#d/dx(f(x))=-(x-9)/(x+3)^3#

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