In this exercise we are going to use the derivative of the quotient#(u(x))/(v(x))# and that of a power function:
#color(red)(((u(x))/(v(x)))'=(u'(x)v(x)-v'(x)u(x))/(v(x))^2#
#color(blue)(((u(x))^n)'=n*u(x)*u'(x))#
We will apply the following polynomial identities:
#color(brown)((a+b)^2=a^2+2ab+b^2)#
#color(purple)(a^2-b^2=(a-b)(a+b))#
Based on the above properties let us compute #f'(x)#
#f'(x)=color(red)((x'*(x+3)^2-color(blue)(((x+3)^2)')*x)/((x+3)^2)^2#
#f'(x)=((x+3)^2-color(blue)(2(x+3)^(2-1)*(x+3)')*x)/(x+3)^4#
#f'(x)=(color(brown)(x^2+6x+9)-2(x+3)(1)*x)/(x+3)^4#
#f'(x)=(x^2+6x+9-2x(x+3))/(x+3)^4#
#f'(x)=(x^2+6x+9-2x^2-6x)/(x+3)^4#
#f'(x)=(-x^2+9)/(x+3)^4#
#f'(x)=color(purple)(9-x^2)/(x+3)^4#
#f'(x)=color(purple)(3^2-x^2)/(x+3)^4#
#f'(x)=color(purple)((3-x)(3+x))/(x+3)^4#
Simplify the fraction by #x+3#
#f'(x)=color(purple)((3-x)cancel((3+x)))/(x+3)^(cancel(4) color(orange)3 )#
#f'(x)=(3-x)/(x+3)^3#
