I know have the final answer and know I need to use the natural log but I'm confused about why that is.

Could someone walk through it step by step?

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    $\begingroup$ $f(x)^{g(x)}=(e^{\ln f(x)})^{g(x)}=e^{g(x)\ln f(x)}$ $\endgroup$ – anon Jun 18 '15 at 1:57

Let $y= \sqrt{x}^{\sqrt x}$. Then $\ln(y) = \sqrt{x} \ln(\sqrt x) = \frac{1}{2} \sqrt x \ln(x)$. So, $$ \frac{d}{dx} \ln(y) = \frac{d}{dx} \left(\frac{1}{2}\sqrt x\ln(x)\right) $$ $$ \frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right) $$ So, $$ \frac{dy}{dx} = y \cdot \frac{1}{2} \left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right) = \frac{1}{2}\sqrt{x}^{\sqrt x}\left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right) $$


The idea is to turn the problem into something to which one can apply the standard rules ( product and chain rules, for instance) to functions the derivative of which one already knows. Now, in your case you have a function raised to another function - which is complicated: there is no 'exponentiation rule' for derivatives, as such (i.e., both the base and exponent are not constants - you know how to take the derivative of $x^n$ and $ e^x $ - but not of $x^x$, at least without some manipulation); but $$ \ln ( a ^ b ) = b \ln a $$ i.e., if you introduce a logarithm into the equation, the exponentiation becomes multiplication - so we can hope to use the product rule.

Now, you do not have a $\ln$ in the question - if you introduce one, you must add at the same time, its inverse, an exponentiation:

$$ a^b = e^{\ln a^b } = e ^ { b \ln a }$$

While this might look like we are complicating things, we are not: we know how to take the derivative of "e to the something" and the derivative of the "ln of something" - and the weird exponentiation has become a product, for which there is a differentiation rule...

So - write

$$ (\sqrt x )^{\sqrt x} = e ^ { \ln ( { \sqrt x } ^ { \sqrt x } )} = e^ { \sqrt x \ln ( \sqrt x) } $$ - the derivative of the function on the right will be the derivative of the function on the left (i.e., of the one you want).

Now, the derivative of $$ e^{whatever}$$ is, by the chain rule, $$ e ^ { whatever } ( derivative\ of\ whatever ). $$ Here 'whatever' is $$ \sqrt x \ln ( \sqrt x )$$ Are you OK, at this point - or do you need more help?

*Edit - For completeness sake, although at the risk of making this long answer too long... Other answers here (contrary, perhaps, to appearances) are doing the same thing as this one; the point is to use a logarithm to turn an exponentiation into a multiplication: if $$ y = {f(x)}^{g(x)},$$ then $$ \ln y = \ln \left({f(x)}^{g(x)}\right) = g(x) \ln f ( x ) ,$$ i.e., the original exponentiation on the right of the equal side becomes a product - so the product rule now applies.

In your case, $g(x) = f(x) = \sqrt x$... Be that as it may, $g(x) \ln ( f(x) )$ is what was called 'whatever' above.

Taking the derivative on both sides of the equation - and remembering the chain rule - one gets (writing it to make it look like the first version in this answer) $$ {y'\over y} = derivative\ of\ whatever.$$ Cross-multiplying with $y$, one obtains

$$ y' = y \ \cdot \ (derivative\ of\ whatever),$$ as was the case in the 'logarithmic derivative' answers - but also in this answer! To be explicit, with your $y$, $$ y = {\sqrt x}^{\sqrt x} = e^{\sqrt x \ln \sqrt x }=e^{whatever},$$ so the methods are the same.

To summarize: (either version of) the method is useful if there is an exponentiation with both base and exponent not constant.

  • $\begingroup$ You have the first identity wrong: $a^b=e^{b \ln(a)}$ $\endgroup$ – ASKASK Jun 18 '15 at 2:25
  • $\begingroup$ that's embarrassing! Thanks! $\endgroup$ – peter a g Jun 18 '15 at 2:26
  • $\begingroup$ No problem! Thankfully it had no effect on the rest since in this case, $a=b=\sqrt{x}$ $\endgroup$ – ASKASK Jun 18 '15 at 2:28
  • $\begingroup$ Well - it rather detracted from the pedagogical value. $\endgroup$ – peter a g Jun 18 '15 at 2:33

let $y=\sqrt{x}^\sqrt{x}$, then $\ln{y}=\sqrt{x}\ln{\sqrt{x}}=\frac{1}{2}x^{\frac{1}{2}}\ln{x}$. Then take $d(.)$ on both sides, we have:

$$ \frac{dy}{y}=\frac{1}{2}(\frac{1}{2}x^{-\frac{1}{2}}\ln{x}+x^{\frac{1}{2}}\frac{1}{x})dx $$

$$ \frac{dy}{dx}=\frac{y}{2}x^{-\frac{1}{2}}(1+\frac{1}{2}\ln{x})=\frac{x^{\sqrt{x}-1}}{2}(1+\frac{1}{2}\ln{x}) $$

  • $\begingroup$ misprint - you lost one of your $1/2$'s on going from the first line to the second line - compare your answer with Andrew's above... $\endgroup$ – peter a g Jun 19 '15 at 12:52
  • $\begingroup$ @peterag, thanks, edited. $\endgroup$ – Indominus Jun 19 '15 at 12:57
  • $\begingroup$ sure - it wasn't as embarrassing or disastrous as my insane misprint that ASKASK noted in my answer... $\endgroup$ – peter a g Jun 19 '15 at 13:00

Notice that $\sqrt{x}^{\sqrt{x}}=x^{\frac{1}{2}\sqrt{x}}$. Applying the chain rule yields the derivative: $$\left(\frac{1}{2}\sqrt{x}\right)\cdot x^{\frac{1}{2}\sqrt{x}-1}+(\ln x)\cdot \frac{1}{2}\cdot \frac{1}{2\sqrt{x}}\cdot x^{\frac{1}{2}\sqrt{x}}.$$

  • $\begingroup$ The chain rule is - of course - $ (f \circ g )' (x) = f'( g(x) ) g'(x)$: what are your $f$ and $g$ here? $\endgroup$ – peter a g Jun 19 '15 at 12:42
  • $\begingroup$ Formally, consider the function $f(y,z)=y^z$. Apply the chain rule where $y=y(x)$ and $z=z(x)$. (It's not the 1 variable chain rule...) $\endgroup$ – pre-kidney Jun 19 '15 at 14:10
  • $\begingroup$ OK! Good! Chances are, though, that the questioner didn't know it... You might want to write it out explicitly in your answer - if you think it's worth it. $\endgroup$ – peter a g Jun 19 '15 at 14:54

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