Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The answer is (I think) $x^{\sqrt{x}-0.5} (1+0.5\ln(x))$, but how?

share|cite|improve this question
Hint: See if you can follow this one Derivative of $x^{x}$. Can you emulate it? – Amzoti Dec 19 '12 at 21:05
Hint: Write $x^{x^{1/2}} = e^{\ln(x)\cdot x^{1/2}}$ – andybenji Dec 20 '12 at 2:55

4 Answers 4

up vote 17 down vote accepted

The way to differentiate most expressions of the form $x^{f(x)}$ is to rewrite them as $e^{f(x)\ln(x)},$ and then use the chain and product rules, and that works in this case too.

share|cite|improve this answer

I'll probably get stones thrown at me for suggesting this, but let's view $x^{f(x)}$ as a hybrid of a power function $x^k$ and a composite exponential function $b^{f(x)}$. It's actually legitimate to take derivatives of these two simpler functions and add them together to get the derivative. $$\begin{align}\frac{d}{dx}x^{f(x)}&=\overbrace{f(x)x^{f(x)-1}}^{\text{power rule}}+\overbrace{x^{f(x)}\ln(x)}^{\text{exponential rule}}\cdot\overbrace{f'(x)}^{\text{chain rule}}\\ &=x^{f(x)} \left(\frac{f(x)}{x}+\ln(x)f'(x)\right)\end{align}$$ which is completely consistent with the "correct" approach of rewriting the function as $e^{f(x)\ln(x)}$ and using the chain and product rules. I don't recommend that you think this way, I just find this curious. I'd love to see a rigorous explanation for this happy coincidence that does not essentially appeal to rewriting as $e^{f(x)\ln(x)}$.

I think I found one. Let $F(y,z)=y^z$ with $y=g(x)$ and $z=f(x)$, then actually $F(y(x),z(x))=g(x)^{f(x)}$. Now using the chain rule for multivariate functions, $$\begin{align}\frac{d}{dx}g(x)^{f(x)}=\frac{dF}{dx}&=\frac{\partial F}{\partial y}\frac{dy}{dx}+\frac{\partial F}{\partial z}\frac{dz}{dx}\\&=zy^{z-1}\frac{dy}{dx}+y^z\ln(y)\frac{dz}{dx}\\&=f(x)(g(x))^{f(x)-1}g'(x)+g(x)^{f(x)}\ln(g(x))f'(x)\\&=g(x)^{f(x)}\left(\frac{f(x)g'(x)}{g(x)}+\ln(g(x))f'(x)\right)\end{align}$$

share|cite|improve this answer
There is some discussion of this point at – MJD Feb 15 '14 at 17:24

Let's write $E'$ to mean the derivative of the expression $E$.

Then in general: $$\begin{align} u^v & = e^{v\log u} \\ (u^v)' &= \left(e^{v\log u}\right)'\\ &=e^{v\log u}\cdot (v\log u)'\\ &=u^v \left(v'\log u + v\frac {u'}u\right) \\ &=u^vv'\log u + u^{v-1}vu' \end{align}$$

To check, try taking one of $u$ or $v$ to be the identity function and the other to be a constant.

share|cite|improve this answer

$$y=x^{\sqrt{x}}$$ $$\ln y=\sqrt{x}\ln x$$ now differentiate w.r.t. x using chain rule $$\dfrac {1}{y}\cdot\dfrac {dy}{dx}=\dfrac{\sqrt{x}}{x}+\dfrac{\ln x}{2\sqrt x}$$ $$\dfrac {dy}{dx}=x^{\sqrt{x}}\cdot\left(\dfrac{2+\ln x}{2\sqrt x}\right)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.