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is there any rule to differentiate the function $(a\,x)^{b\,x}$?

I've got to find the derivative of $(x^2+1)^{\arctan x}$ and Wolfram|Alpha says the answer is $$\tan^{-1}(x) (x^2+1)^{\tan^{-1}(x)-1} \left(\frac{d}{dx}(x^2+1)\right)+\log(x^2+1) (x^2+1)^{\tan^{-1}(x)} \left(\frac{d}{dx}(\tan^{-1}(x))\right)$$ Is there any general rule to do that? Thanks.

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up vote 4 down vote accepted

Your derivative of $(x^2+1)^{\arctan x}$ is the particular case for $u(x)=x^2+1$ and $v(x)=\arctan x$ of

$$\frac{d}{dx}\left(\left[ u(x)\right] ^{v(x)}\right)=v(x)\left[ u(x)\right] ^{v(x)-1}u^{\prime }(x)+\left( \ln u(x)\right) \left[ u(x)\right] ^{v(x)}v^{\prime }(x),$$

or omitting de variable $x$:

$$\left( u^{v}\right)^{\prime }=vu^{v-1}u^{\prime }+\left( \ln u\right) u^{v}v^{\prime }.$$

This can be derived observing that, since $u=e^{\ln u}$, we have $u^v=e^{v\ln u}$:

$$\begin{eqnarray*} \frac{d}{dx}\left( u^{v}\right) &=&\frac{d}{dx}\left( e^{v\ln u}\right) \\ &=&e^{v\ln u}\left( \ln u\frac{dv}{dx}+\frac{v}{u}\frac{du}{dx}\right) \\ &=&u^{v}\left( \ln u\frac{dv}{dx}+\frac{v}{u}\frac{du}{dx}\right) \\ &=&u^{v}\ln u\frac{dv}{dx}+u^{v}\frac{v}{u}\frac{du}{dx} \\ &=&u^{v}(\ln u)v'+u^{v-1}vu'. \end{eqnarray*}$$

Another possibility is to consider the variables $u$ and $v$ (both depending on $x$) in the function

$$z=f(u(x),v(x))=\left[ u(x)\right] ^{v(x)}$$

and determine its total derivative with respect to $x$:

$$\begin{eqnarray*} \frac{dz}{dx} &=&\frac{d}{dx}\left( \left[ u(x)\right] ^{v(x)}\right) \\ &=&\frac{% \partial z}{\partial u}\frac{du}{dx}+\frac{\partial z}{\partial v}\frac{dv}{% dx} \\ &=&v(x)\left[ u(x)\right] ^{v(x)-1}u^{\prime }(x)+\left[ u(x)\right] ^{v(x)}\left( \ln u(x)\right) v^{\prime }(x) \end{eqnarray*}$$


$$\frac{\partial z}{\partial u}=\frac{\partial }{\partial u}\left( u^{v}\right) =vu^{v-1}$$


$$\frac{\partial z}{\partial v}=\frac{\partial }{\partial v}\left( u^{v}\right) =\frac{\partial }{\partial v}\left( e^{\ln u\cdot v}\right) =e^{\ln u\cdot v}\ln u=u^{v}\ln u.$$

For $u(x)=ax,v(x)=bx$, we get

$$\frac{d}{dx}\left( \left( ax\right) ^{bx}\right) =bx\left( ax\right) ^{bx-1}a+\left( ax\right) ^{bx}\left( \ln (ax)\right) b.$$

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Assuming you mean $(ax)^{bx}$, I'd just write it as $(e^{\ln(ax)})^{bx}$ and use the chain rule (ie $e^{\ln(ax)bx} = e^{u(x)}$ and go from there).

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HINT $\rm\ \ (F^G)'\ =\ (e^{G\:ln\ F})'\: =\ F^G\ (GF'/F + G'\ ln\ F)$

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I think there is a typo in the sign. – Américo Tavares Mar 6 '11 at 16:26
@Americo: Fixed it, thanks! – Bill Dubuque Mar 6 '11 at 16:32

Not answering the math part, since Stijn has already done that, but if you click on the "show steps" button, Wolfram|Alpha shows you one possible path for the derivation. I've included the image for the derivation of $\frac{\partial}{\partial x} ((a x)^{b x})$ enter image description here

A similar set of steps is supplied for the other derivative that you want to take:^2%2B1)^(tan^(-1)(x)))

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The link for the first example is[(ax)^(bx),x]. Does anyone know how to get Wolfram|Alpha links working properly in the markup? – Simon Mar 6 '11 at 11:41
The parentheses and the caret confuse Markdown. Try escaping ^ with %5E, ( with %28, ) with %29, [ with %5B and ] with %5D like this: WA Link (you obviously have to use the [title](URL) syntax) – kahen Mar 6 '11 at 13:21

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