How do you differentiate # f(x)=sqrt([(2x-5)^5]/[(x^2 +2)^2] # using the chain rule.?

1 Answer
Jan 28, 2017

Answer:

#f'(x)=((2x-5)^2(x^2+10x+10))/((x^2+2)^2sqrt(2x-5))#

Explanation:

#f'(x)=1 /(2sqrt((2x-5)^5/(x^2+2)^2))color(red)(* (5*(2x-5)^4*2*(x^2+2)^2-(2x-5)^5*2*(x^2+2)*2x)/(x^2+2)^4)#

#=1/2sqrt(((x^2+2)^2)/(2x-5)^5 ) * ( 10*(2x-5)^4*(x^2+2)^2-4x*(2x-5)^5*(x^2+2))/(x^2+2)^4#

#=1/2sqrt(((x^2+2)^2)/(2x-5)^5)*(2 (2x-5)^4 (x^2+2)(5(x^2+2)-2x(2x-5)))/(x^2+2)^4#

#=1/cancel2sqrt(((x^2+2)^2)/(2x-5)^5)*(cancel2 (2x-5)^4 cancel((x^2+2))(5x^2+10-4x^2+10x))/(x^2+2)^(cancel4^3#

#=sqrt(((x^2+2)^2)/(2x-5)^5)*((2x-5)^4(x^2+10x+10))/(x^2+2)^3#

#=cancel((x^2+2))/(cancel((2x-5)^2)sqrt(2x-5))*((2x-5)^(cancel4^2)(x^2+10x+10))/(x^2+2)^(cancel3^2#

#=((2x-5)^2(x^2+10x+10))/((x^2+2)^2sqrt(2x-5))#