How do you differentiate # f(x)=x/sqrt(3-xe^x)# using the chain rule.?

1 Answer
Jan 15, 2016

Answer:

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

Explanation:

#f(x)=x/sqrt(3-xe^x)=sqrt(x^2/(3-xe^x))#

Using the Chain Rule

given

#y=f(g(x))#

we could think

#u=g(x)#; #y=f(u)#

#dy/dxf(g(x))=d/(du)f(u)*(du)/dx=f'(u)*u'=#

#=f'(g(x))*g'(x)#

There are two ways we could follow:

  • We consider #y=[f(g(x))]^n#

#f'(x)=n[f(g(x))]^(n-1)*d/(dx)g(x)=#

#=1/2(x^2/(3-xe^x))^(1/2-1)d/(dx)(x^2/(3-xe^x))=#

#=1/2(x^2/(3-xe^x))^(-1/2)(2x(3-xe^x)-x^2(0-(e^x+xe^x))/(3-xe^x)^2)=#

#=1/(2sqrt(x^2/(3-xe^x)))*(6-xe^x+x^2e^x)/((3-xe^x)^2)*x=#

#=1/2(sqrt(3-xe^x)/color(green)cancel(x))(6-xe^x+x^2e^x)/((3-xe^x)^2)*color(green)cancel(x)=#

#=1/2sqrt(color(magenta)cancel((3-xe^x))/(3-xe^x)^(color(magenta)(cancel(4)^3)))(x^2e^x-xe^x+6)=#

#=1/(2sqrt((3-xe^x)^3))(x^2e^x-xe^x+6)#

  • We consider #y=sqrt(g(x))#

#f'(x)=1/(2sqrt(g(x)))d/(dx)g(x)=#

#=1/(2sqrt(x^2/(3-xe^x)))*((2x(3-xe^x)-x^2(0-e^x+e^x(-x)))/(3-xe^x)^2)=#

#=1/2(sqrt((3-xe^x)/x^2))*(6x-2x^2e^x+x^2e^x+x^3e^x)/((3-xe^x)^2)=#

#=1/(2color(green)cancel(x))*sqrt(3-xe^x)((color(green)cancel(x)(6-xe^x+x^2e^x))/((3-xe^x)^2))=#

#=1/2sqrt(color(magenta)cancel((3-xe^x))/(3-xe^x)^(color(magenta)(cancel(4)^3)))(x^2e^x-xe^x+6)=#

#=1/(2sqrt((3-xe^x)^3))(x^2e^x-xe^x+6)#