How do you differentiate #g(x) = (1/x^3)*sqrt(x-e^(2x))# using the product rule?

Question

1726 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-22T08:51:17

#-3x^-4##sqrt(x-e^(2x)) +x^-3 1/2(x-e^(2x))^(-1/2)(1-2e^(2x))#

#f(x)g(x)=d/dxf(x)g(x)+f(x)d/dxg(x)#

So, we have #(1/x^3)=x^-3#

#d/dx x^-3=-3x^-4#using this rule #nx^(n-1)#

#-3x^-4##sqrt(x-e^(2x))+x^-3#

then the derivative of #sqrt(x-e^(2x)##=##(x-e^(2x))^(1/2)#

#(x-e^(2x))^(1/2)# using chain rule enter image source here

#(x-e^(2x))^(1/2)#= 1/2(x-e^(2x))^(-1/2)(1-2e^(2x))#

So, it's equal to
#-3x^-4##sqrt(x-e^(2x)) +x^-3 1/2(x-e^(2x))^(-1/2)(1-2e^(2x))#