Question

Question #0caf0

Answer 1

`d/dx xln(1/x)=ln(1/x)-1`

Explanation:

Using the product rule

`d/dx f(x)g(x) = f(x)g'(x) + g(x)f'(x)`

along with the derivatives

• `d/dx x = 1`
• `d/dx ln(x) = 1/x`

we have

`d/dx xln(1/x) = d/dx xln(x^(-1))`

`=d/dx (-xln(x))`

`=-x(1/x) + (-1)ln(x)`

`=-1 + ln(x^(-1))`

`=ln(1/x)-1`