# Differentiation (using chain rule?)

I'm looking for a general method to evaluate expressions of the form

$$\frac{\mathrm{d}(u^v)}{\mathrm{d}u}\text{ and }\frac{\mathrm{d}(u^v)}{\mathrm{d}v}\;.$$

I know that the answers to these are, respectively, $u^{v-1}v$ and $u^v\mathrm{ln}u$, but am unsure of how to obtain them, and how the chain rule applies here.

I'd be very grateful of any enlightenment.

With very many thanks,

Froskoy.

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The chain rule doesn't apply here. This is little more than saying $\frac{d}{dx} x^n = nx^{n-1}$ and $\frac{d}{dx} a^x = a^x \ln a$. These are both elementary derivatives, proved with limits. – mixedmath May 31 '12 at 10:09

If $v$ is a constant, $\frac{d}{du}u^v=vu^{v-1}$, but the chain rule is not required. If $v$ is a function of $u$, then your formula is simply wrong: $u^v=\left(e^{\ln u}\right)^v=e^{v\ln u}$, so

$$\frac{d}{du}u^v=\frac{d}{du}e^{v\ln u}=e^{v\ln u}\left(\frac{v}u+\ln u\frac{dv}{du}\right)=u^v\left(\frac{v}u+\ln u\frac{dv}{du}\right)\;,$$ which can be written $$vu^{v-1}+u^n\ln u\frac{dv}{du}\;,$$ if you really wish.

Similarly, if $u$ is constant, $\frac{d}{dv}u^v=u^v\ln u$, no chain rule being required, but if $u$ is a function of $v$, then $$\frac{d}{dv}u^v=v^u\left(\frac{u}v+\ln v\frac{du}{dv}\right)\;.$$

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In the first one, if $v$ is a constant, then the chain rule is not involved. If $v$ is not a constant, but is a function of $u$, then your formula is wrong, and some form of the chain rule is needed. One way to go about it is $u^{v(u)}=e^{v(u)\log u}$; the derivative is $$e^{v(u)\log u}\times{d\over du}(v(u)\log u)=e^{v(u)\log u}\left({dv\over du}\log u+{v(u)\over u}\right)$$

Similar remarks apply to the second one, this time with the question being whether $u$ is a constant or a function of $v$.

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Are you sure about that derivative when $v$ is a function of $u$? – Brian M. Scott May 31 '12 at 10:37
Aargh. Did I get it right this time? – Gerry Myerson May 31 '12 at 11:28