Find the derivative $y=({1\over x})^x$

Question

I got $y'= x^2 + \ln({1\over x})\times ({1\over x})^x$

Am I correct?

So $y'=-x^{-x}(\ln{x} + 1)$

Answer 1

$$y = x^{-x} \implies \log(y) = -x \log(x)$$ Can you now do implicit differentiation?

Answer 2

Some logarithms and the chain rule:

$$y=\left(\frac{1}{x}\right)^x=e^{\frac{1}{x}\log x}\Longrightarrow y'=e^{\frac{1}{x}\log x}\left[-\frac{1}{x^2}\log x+\frac{1}{x^2}\right]=$$

$$=\frac{1}{x^2}\left(\frac{1}{x}\right)^x\left(1-\log x\right)=\frac{1}{x^{x+2}}(1-\log x)$$