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

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Answer 1 · 2016-07-05T16:00:07

#f'(x) = (cosx(1-xe^x) - sinx(-e^x - xe^x))/(1-xe^x)^2#

#d/(dx) ((u(x))/(v(x))) = (u'(x)*v(x) - u(x)*v'(x))/(v(x))^2#

In this case #u(x) = sinx# and #v(x) = 1 - xe^x#

#u'(x) = cosx# but to find #v'(x)# we will also need to use the product rule, which is given by:

#d/(dx) g(x)h(x) = g'(x)h(x) + g(x)h'(x)#

#v'(x) = -e^x - xe^x#

Now we have everything we need, so plug in the functions to obtain

#f'(x) = (cosx(1-xe^x) - sinx(-e^x - xe^x))/(1-xe^x)^2#

This is about as simplified as it's going to get. We can split it into two fractions which could be useful in some scenarios:

#f'(x) = (cosx)/((1-xe^x)) - (sinx(-e^x - xe^x))/(1-xe^x)^2#