Question

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

Answer 1

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

Explanation:

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`