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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!

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Have you studied the technique known as logarithmic differentiation? –  Gerry Myerson Oct 11 '12 at 6:00
    
Yes, I actually just realized what to do, I couldn't see to do logarithmic differentiation because this function was just a part of a larger one, now that I wrote it by itself it makes sense –  Chance Oct 11 '12 at 6:03
    
A function like $f(x)^{x}$ can be rewritten in the form $e^{x\log(f(x))}$, taking care with domain of definition. –  Geoff Robinson Oct 11 '12 at 6:04

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up vote 4 down vote accepted

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 $f\,'(x)$.

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Thanks I figured it out from there. –  Chance Oct 11 '12 at 6:06
    
@Chance: Excellent! –  Brian M. Scott Oct 11 '12 at 6:10

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