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

Question

1519 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-06T23:15:44

#f'(x) = (xcosx - e^xcosx - sinx + e^xsinx)/(x-e^x)^2#

The quotient rule states that:

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

In this case,

#f(x) = sinx" "" "" "=>" "" " f'(x) = cosx#

#g(x) = x-e^x" "" "color(white)"-" => " "" "g'(x) = 1 - e^x#

Therefore, if we differentiate with quotient rule, we get:

#d/dx(sinx/(x-e^x)) = ((x-e^x)(cosx) - (sinx)(1-e^x))/(x-e^x)^2#

#= (xcosx - e^xcosx - sinx + e^xsinx)/(x-e^x)^2#

There are other ways to simplify this, but all forms of it are equally correct.

Final Answer