Take the 2-minute tour ×
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.

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.

share|improve this question

4 Answers 4

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 $x$:

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

because

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

and

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

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

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

share|improve this answer

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).

share|improve this answer

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

share|improve this answer
    
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: http://www.wolframalpha.com/input/?i=d/dx((x^2%2B1)^(tan^(-1)(x)))

share|improve this answer
    
The link for the first example is wolframalpha.com/input/?i=D[(ax)^(bx),x]. Does anyone know how to get Wolfram|Alpha links working properly in the markup? –  Simon Mar 6 '11 at 11:41
1  
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

Your Answer

 
discard

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.