Differentiating $ \left( 1 - \frac {1}{x} \right)^x $ I have a calculus question. How does one differentiate $\left(1-\frac{1}{x}\right)^x$, for x>1? It should be positive right?  
 A: Notice that the power is $x$, which is not a constant, so if we let $ y = \left( 1 - \frac {1}{x} \right)^x $, Then we can take the natural logarithm to get $$ \ln y = \ln \left( 1 - \frac {1}{x} \right)^x = x \cdot \ln \left( 1 - \frac {1}{x} \right). $$
Differentiating each side (implicit differentiation), we get $$ \begin {align*} \frac {y'}{y} &= x \cdot \frac {\left( 1 - \frac {1}{x} \right)'}{1 - \frac {1}{x}} + 1 \cdot \ln \left( 1 - \frac {1}{x} \right) \\&= x \cdot \frac {- \frac {1}{x^2}}{1 - \frac {1}{x}} + \ln \left( 1 - \frac {1}{x} \right) \\&= \frac {1}{1 - x} + \ln \left( 1 - \frac {1}{x} \right), \end {align*} $$so $ y' = \left( 1 - \frac {1}{x} \right)^x \cdot \left( \frac {1}{1-x} + \ln \left( 1 - \frac {1}{x} \right) \right) $. 
Now, I don't know exactly what you're asking since it's unclear what the question is, but hopefully whatever it is, you can finish from here. In general, if $ y = \left[ u(x) \right]^x $, you can follow the same steps and find that $$ y' = \left[ u(x) \right]^{x-1} \cdot x \cdot u'(x) + \left[ u(x) \right]^x \cdot \ln \left( u(x) \right). $$
A: Hint: Let $f(x) = \left(1-\frac{1}{x}\right)^x = \dfrac{(x-1)^x}{x^x} \to \ln f(x) = \ln(x-1)^x -\ln x^x = x\left(\ln(x-1)-\ln x\right)$. Can you take it further?
A: Hint: $\qquad\qquad\bigg[\bigg(1-\dfrac1x\bigg)^x~\bigg]'\quad=\quad\bigg[\bigg(1-\dfrac1x\bigg)^n~\bigg]'_{\large n=x}\quad+\quad\Big(a^x\Big)'_{\large a=1-\frac1x}$
Can you take it from here ? :-)
