How do you differentiate # f(x)=xsqrt(3x-e^x)# using the chain rule.?

1 Answer
May 22, 2018

Answer:

# f'(x)=(9x-xe^x-2e^x)/(2sqrt(3x-e^x))#.

Explanation:

We have, #f(x)=xsqrt(3x-e^x)#.

Using the Product Rule, we get,

#:. f'(x)=x*d/dx{(3x-e^x)^(1/2)}+sqrt(3x-e^x)*d/dx{x}......(star)#.

Here, by the Chain Rule,

#d/dx{(3x-e^x)^(1/2)}=1/2*(3x-e^x)^(1/2-1)*d/dx{(3x-e^x)}#,

#=1/2*(3x-e^x)^(-1/2)*[d/dx{3x}-d/dx{e^x}]#,

#=1/(2sqrt(3x-e^x)){3*1-e^x}#.

#rArrd/dx{(3x-e^x)^(1/2)}=(3-e^x)/(2sqrt(3x-e^x))......(star_1)#.

Utilising #(star_1)" in "(star)#, we have,

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

# ={(3x-xe^x)+2(3x-e^x)}/(2sqrt(3x-e^x))#.

# rArr f'(x)=(9x-xe^x-2e^x)/(2sqrt(3x-e^x))#.

Enjoy Maths.!