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

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Answer 1 · 2015-11-17T08:15:11

Apply the quotient rule to each term to find that #f'(x) = (1-x)/e^x + 1/x^2#

The quotient rule states that
#d/dxf(x)/g(x) = (f'(x)g(x) - f(x)g'(x))/(g(x))^2#

Now, using the following facts:
#d/dx(f(x) + g(x)) = f'(x) + g'(x)#
#d/dx e^x = e^x#
#d/dx x = 1#
#d/dx 1 = 0#

We have
#f'(x) = d/dx(x/e^x - 1/x) = (d/dxx/e^x) + (d/dx-1/x)#

Thus, applying the chain rule twice,

#f'(x) = (e^x - xe^x)/(e^x)^2 + -(0 - 1)/x^2= (e^x(1-x))/(e^x)^2 + 1/x^2#

So, cancelling an #e^x# in the first term, we have

#f'(x) = (1-x)/e^x + 1/x^2#