What is the derivative of $y=x^y$

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What is the derivative of $y=x^y$ ?

I have tried the below. Please correct me if I am wrong on any of the below.

$$y=x^y.$$

Taking natural log on both sides,

$$\log y = y \log x.$$

Differentiating on both sides,

$$\left(\frac1y - \log x\right) y' = \frac{y}x$$

$$y' = \frac{y}{(\frac{x}y) - x\log x}$$

$$y' = \frac{y^2}{x(1-y\log x)}$$

Finally,

$$y' = \frac{x^{2y}}{x(1-x^y\log x )}$$

$$y' = \frac{x^{2y-1}}{1-x^y \log x}.$$

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2 Answers

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$$y=e^{y\ln (x)} $$

$$\ln (y)=y\ln (x) $$

$$x=e^{\frac {\ln (y)}{y}} $$

differentiating wrt $x $, we find

$$1=x \frac {1-\ln (y)}{y^2}y'$$

thus

$$y'=\frac {y^2}{x (1-\ln (y))} $$

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$$y'=(x^y)'=\left(e^{y\ln{x}}\right)'=x^y\left(y'\ln{x}+\frac{y}{x}\right),$$ which gives $$y'=\frac{yx^{y-1}}{1-x^y\ln{x}}$$

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