Derivative of a function raised to the power of x I have come across a derivative that I don't know how to solve. 
$$(x^{\frac{1}{2}} + \ln x)^x.$$
I know how to take derivatives of a constant to the power of $x$ but for some reason I can't figure this one out, any help would be great!
 A: You want the notion of a logarithmic derivative. I’ll give you the general principle; if you still can’t work it out, let me know, and I’ll apply it to your problem.
You have a function $f(x)=u(x)^{v(x)}$, and you want $f\,'(x)$. Start by taking logs to get $\ln f(x)=\ln u(x)^{v(x)}=v(x)\ln u(x)$. Now differentiate:
$$\frac{f\,'(x)}{f(x)}=v(x)\cdot\frac{u'(x)}{u(x)}+v'(x)\ln u(x)\;,$$
Finally, multiply both sides by $f(x)$, and you’ll have $f\,'(x)$.
A: Such type of confusing functions can be differentiated easily by following steps:


*

*Parameterize the confusing terms in the given function. In the given example, function of $x^x$, we can substitute for $x$ in power as $y$. Then the given function actually becomes a function of two variables $x$ and $y$.

*The function obtained after step 1 above can be easily differentiated by partial differentiation method.

*In the differentiated function obtained after step 2, substitute $y=x$ and you will have the differentiated function of originally considered confusing function in $x$.
The image of calculations for given function based on above steps is uploaded here. enter image description here
A: Alternatively, you can express the function as follows: $f(x)=e^{\ln{f(x)}}$, whose derivative is:
$$f'(x)=e^{\ln{f(x)}}\cdot (\ln{f(x)})'.$$
Thus, for $f(x)=(x^{1/2}+\ln{x})^x$:
$$f'(x)=e^{\ln{(x^{1/2}+\ln{x})^x}}\cdot (\ln{(x^{1/2}+\ln{x})^x})'=$$
$$(x^{1/2}+\ln{x})^x\cdot(x\cdot\ln{(x^{1/2}+\ln{x})})'=$$
$$(x^{1/2}+\ln{x})^x\cdot\left(\ln{(x^{1/2}+\ln{x})}+x\cdot \frac{1}{x^{1/2}+\ln x}\cdot(\frac{1}{2x^{1/2}}+\frac{1}{x})\right).$$
Note: This method is equivalent to the method given by Brian M. Scott. However, it allows one to keep everything on one side.
edit: fixed a derivative
