How do you differentiate #f(x)=sqrt((1-xe^(2x))^3# using the chain rule.?

1 Answer
Oct 11, 2016

Answer:

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

Explanation:

#f(x)=sqrt((1-xe^(2x))^3)#

#[1]" "D_x[sqrt((1-xe^(2x))^3)]#

The derivative of #sqrt(x)# is #1/(2sqrt(x))#

#[2]" "=1/(2sqrt((1-xe^(2x))^3))*D_x[(1-xe^(2x))^3]#

Use power rule

#[3]" "=1/(2sqrt((1-xe^(2x))^3))*3(1-xe^(2x))^2*D_x[1-xe^(2x)]#

#[4]" "=(3(1-xe^(2x))^2)/(2sqrt((1-xe^(2x))^3))*[D_x[1]-D_x[xe^(2x)]]#

The derivative of a constant is 0

#[5]" "=(3(1-xe^(2x))^2)/(2sqrt((1-xe^(2x))^3))*[-D_x[xe^(2x)]]#

#[6]" "=(-3(1-xe^(2x))^2)/(2sqrt((1-xe^(2x))^3))*[D_x[xe^(2x)]]#

Use product rule

#[7]" "=(-3(1-xe^(2x))^2)/(2sqrt((1-xe^(2x))^3))*[D_x[x]*e^(2x)+x*D_x[e^(2x)]]#

The derivative of #x# is #1#

#[8]" "=(-3(1-xe^(2x))^2)/(2sqrt((1-xe^(2x))^3))*[e^(2x)+x*D_x[e^(2x)]]#

You can think of #e^(2x)# as #(e^x)^2#

#[9]" "=(-3(1-xe^(2x))^2)/(2sqrt((1-xe^(2x))^3))*[e^(2x)+x*D_x[(e^x)^2]]#

Use power rule

#[10]" "=(-3(1-xe^(2x))^2)/(2sqrt((1-xe^(2x))^3))*[e^(2x)+x*2e^x*D_x[(e^x)]]#

The derivative of #e^x# is also #e^x#

#[11]" "=(-3(1-xe^(2x))^2)/(2sqrt((1-xe^(2x))^3))*(e^(2x)+x*2e^x*e^x)#

#[12]" "=color(red)((-3(1-xe^(2x))^2)/(2sqrt((1-xe^(2x))^3))*(e^(2x)+2xe^(2x)))#

You could leave it there or you can simplify it further if you are required to.

#[13]" "=(-3(1-xe^(2x))^2)/(2(1-xe^(2x))^(3/2))*(e^(2x)+2xe^(2x))#

#[14]" "=-3/2(1-xe^(2x))^(1/2)*(e^(2x)+2xe^(2x))#

#[15]" "=color(red)(-3/2sqrt(1-xe^(2x))e^(2x)(1+2x))#