Can someone help to differentiate this function $1/x^x$?

Question

My teacher let me investigate how to differentiate $${1\over x^x}$$ but I don't really know how to do it. I tried $$x^n = nx^{n-1}$$ and L'Hopital's rule but that did not work or maybe I didn't do correctly. If someone could help me I would be really grateful :D

Answer 1

You have to consider that :

$$\frac1{x^x}=e^{-x\ln(x)}$$

This allows to compute the derivative :

$$\frac{d}{dx}\left(\frac1{x^x}\right)=-\left(1+\ln(x)\right)e^{-x\ln(x)}$$

Answer 2

Rewrite the function as $y=x^{-x}$ the best way to do this to make sure you don't forget your chain rule is to use logarithmic differentiation. $\ln y=-x\ln x\to \frac{y\prime}{y}=-\ln x-1\to y\prime(-\ln x-1)=-x^{-x}(\ln x+1)$

Answer 3

Note that: $$x^x=e^{x\ln{x}}$$ Hence, you must differentiate: $$\frac{d}{dx}\left(\frac{1}{x^x}\right)=\frac{d}{dx}(e^{-x\ln{x}})$$ Now, you can use a combination of the product rule and chain rule.

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and L'Hopital's rule but that did not work or maybe I didn't do correctly

Note that you cannot use L'Hopital's rule to evaluate derivatives. However, it is useful when evaluating certain limits. For derivatives, something similar would be the quotient rule, however note that you'd have evaluate the derivative of $x^x$ first.

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I tried $x^n=nx^{n-1}$

Also, note that you cannot use the power rule in this case: $$\frac{d}{dx}(x^n)=nx^{n-1}$$ Since that only works if $n$ is a real valued constant. In this case, we are varying $x$, thus we cannot use it.

Answer 4

By the multivariable chain rule:

Let $y=u^v$, $u=x$, and $v=-x$. Then

$$\begin{align}\frac{dy}{dx}&=\frac{\partial y}{\partial u}\frac{du}{dx}+\frac{\partial y}{\partial v}\frac{dv}{dx}\\&=vu^{v-1}-\ln(u)u^v\\&=-(1+\ln(x))x^{-x}\end{align}$$