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Find the derivative of $y=x^x$.

My Attempt: $$y=x^x$$ Taking $\textrm {ln}$ on both sides, we get: $$\textrm {ln} y= \textrm {ln} x^x$$ $$\textrm {ln} y = x \textrm {ln} x$$

How do I procees further?

pi-π
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  • Use the chain rule on $\ln(y)$: the derivative of the inner function is simply $y'$. – Chris May 18 '17 at 15:33
  • Now implicitly differentiate both sides with respect to x. – Tucker May 18 '17 at 15:34
  • after your last step, differentiate both sides wrt $x$, then you are at the answer – Hirakjyoti Das May 18 '17 at 15:34
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    Or think how you might calculate it $y = e^{x\cdot\log(x)}$. Now apply a couple of common rules. – badjohn May 18 '17 at 15:35
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    One might think that $(x^x)' = x \cdot x^{x-1}$, since that works for constant powers, but that's not quite right. One might also think that $(x^x)' = \ln(x) x^x$ since that's how it works for constant bases, but that's not quite right either. Amazingly adding these together gives the correct answer: $(x^x)' = x \cdot x^{x-1} + \ln(x) x^x$. – Alice Ryhl May 18 '17 at 15:37
  • @AliceRyhl That's the chain rule for $y=y(s(x),t(x))$. – Zain Patel May 18 '17 at 15:45
  • I suppose the question implicitely implies that $x>0$, since studying the (not defined everywhere) function $y=x^x$ when $x<0$ is a much more difficult task... – Evargalo May 18 '17 at 16:03

5 Answers5

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Here is an alternative way to do it, without the use of implicit differentiation:

One can start by writing: $$y=x^x=(e^{\ln{x}})^x=e^{x\ln{x}}$$ Hence, it now becomes extremely obvious to apply the chain rule to obtain: $$\frac{dy}{dx}=\frac{d}{dx}(e^{x\ln{x}})=\color{green}{\frac{d}{dx}(x\ln{x})}\cdot e^{x\ln{x}}$$ Now all you need to do is apply the product rule on the green one.

projectilemotion
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your start is very good: $$\frac{1}{y}y'=\ln(x)+x\frac{1}{x}$$

Dr. Sonnhard Graubner
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Differentiate with respect to $x$ and on right hand side use product rule.

$\frac 1y \frac{dy}{dx} = \frac{d}{dx} x \cdot \ln x + x \cdot \frac{d}{dx} \ln x$

$\frac 1y \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac 1x$

$\frac 1y \frac{dy}{dx} = \ln x + 1$

$\frac{dy}{dx} = y(\ln x + 1)$

$\frac{dy}{dx} = x^x(\ln x + 1)$

Kanwaljit Singh
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Despite being an answer

Instead of "How do I proceed further?" one may ask as well "How can I preprocess?".

Here's a proposal: Searching for $\,$ x^x $\,$ in $\,$ https://approach0.xyz/search/ $\,$ yields many$^\text{many}$ hits. Ordered by age – or ripeness? – you may choose & click

Worthwhile & close to it is Is $x^x$ an exponential function? (just having received an upvote).

Still link no. 7
How effectively to search through MSE resources
before posting this one.

Hanno
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Remark that $y = x^x=e^{x \log x}$ and then proceed with the product rule $\frac{dy}{dx} = e^{x \log x} \frac{d}{dx} (x \log x) = x^x (\log x + 1)$.

Hasek
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