How do you differentiate #f(x)= 1/(e^(3x) -x)# using the quotient rule?

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Answer 1 · 2018-04-02T19:11:40

The derivative is #-(3e^(3x)-1)/[e^(3x)-x]^2#.

#d/dx(f(x)/g(x))quad=quad(f'(x)g(x)-f(x)g'(x))/[g(x)]^2#

Let's use this in our problem:

#color(white)=d/dx(1/(e^(3x)-x))#

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

#=(color(red)cancelcolor(black)(0*(e^(3x)-x))-1*d/dx(e^(3x)-x))/[e^(3x)-x]^2#

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

#=-(d/dx(e^(3x))-d/dx(x))/[e^(3x)-x]^2#

#=-(3e^(3x)-d/dx(x))/[e^(3x)-x]^2#

#=-(3e^(3x)-1)/[e^(3x)-x]^2#

That's it. Hope this helped!