How do you integrate #y=(5e^-x)/(x+e^(-2x))# using the quotient rule?

Question

1399 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-19T02:22:37

#dy/dx=(-5e^-x(x+1)+5e^(-3x))/(x+e^(-2x))^2#

I think you meant "how do you differentiate using the quotient rule".

If #y=f(x)/g(x)# then #dy/dx=(g(x)*f'(x)-f(x)*g'(x))/(g(x))^2#

Our #f(x)=5e^-x# and #g(x)=x+e^(-2x)#. Then:

#f'(x)=-5e^-x# and #g'(x)=1-2e^(-2x)#

Now let's plug them in:

#dy/dx=((x+e^(-2x))(-5e^-x)-(5e^-x)(1-2e^(-2x)))/(x+e^(-2x))^2#

#dy/dx=(-5xe^-x-5e^(-3x)-5e^-x+10e^(-3x))/(x+e^(-2x))^2#

#dy/dx=(-5e^-x(x+1)+5e^(-3x))/(x+e^(-2x))^2#