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I am suppose to differentiate

$y=(\sin x)^{\ln x}$

I have absolutely no idea, this was asked on a test and I just do not know how to do this I have forgotten the tricks I was suppose to memorize for the test.

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If you take the logarithms of both sides, you have $\ln\,y=(\ln\,x)(\ln\sin\,x)$. You can use implicit differentiation, the chain rule, and the product rule from here. –  J. M. May 4 '12 at 14:48
    
I don't have those things memorized anymore. Also I don't know where the lnx went. –  user138246 May 4 '12 at 14:51
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You may not have them memorized, but you can look them up: implicit differentiation, the chain rule and the product rule. –  Chris Taylor May 4 '12 at 14:59
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Unsolicited advice: "memorizing" everything, and "reviewing without looking things up" are generally bad study policies. –  rschwieb May 4 '12 at 15:03
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@Jordan $y=(\sin x)^{\ln x}$ means that $y$ is $\sin x$ raised to the $\ln x$ power, while $\sin x^{\ln x}=\sin (x^{\ln x})$. –  Américo Tavares May 4 '12 at 16:52
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2 Answers

up vote 5 down vote accepted

Hint

$$\sin x=e^{\ln \left( \sin x\right) }\Rightarrow \left( \sin x\right) ^{\ln x}=\left( e^{\ln \left( \sin x\right) }\right) ^{\ln x}=e^{\left( \ln x\right) \;\cdot\;\ln \left( \sin x\right) }\tag{1}$$

and evaluate the derivative of $e^{\left( \ln x\right) \ln \left( \sin x\right) }.$

Comments (trying to reply to OP's comments).

  1. We can start by writing the given function as $$y=(\sin x)^{\ln x}=e^{\left( \ln x\right) \;\cdot\;\ln \left( \sin x\right)},\tag{2}$$ which is a particular case of the algebraic identity $$\left[ u\left( x\right) \right] ^{v\left( x\right) }=e^{v(x)\;\cdot\;\ln(u(x))}.\tag{3}$$ Remarks. We've used the following properties. By the definition of the natural logarithm, we have (see Powers via logarithms) $$\ln u=v\Leftrightarrow u=e^v=e^{\ln u},\tag{4}$$ and the rule $(a^b)^c=a^{b\;\cdot\; c}\tag{5}$
  2. Finally we evaluate the derivative of $e^{g(x)}$, where $$g(x)=\left( \ln x\right)\;\cdot\; \ln \left( \sin x\right)\tag{6}.$$ By the chain rule we have $$y'=(e^{g(x)})'=e^{g(x)}g^{\prime }(x),\tag{7}$$ and $g'(x)$ is to be computed by the product rule.

Evaluation of $(7)$ $$\begin{eqnarray*} g^{\prime }(x) &=&\left( \left( \ln x\right) \ln \left( \sin x\right) \right) ^{\prime } \\ &=&\left( \ln x\right) ^{\prime }\ln \left( \sin x\right) +\left( \ln x\right) \left( \ln \left( \sin x\right) \right) ^{\prime } \\ &=&\frac{1}{x}\ln \left( \sin x\right) +\left( \ln x\right) \frac{\cos x}{ \sin x} \\ &=&\frac{\ln \left( \sin x\right) \sin x+\left( \left( \ln x\right) \cos x\right) x}{x\sin x}.\tag{8} \end{eqnarray*}$$

Hence, since $e^{g(x)}=\left( \sin x\right) ^{\ln x}$, we obtain $$\begin{eqnarray*} y^{\prime } &=&e^{g(x)}g^{\prime }(x)=y\;\cdot\; g^{\prime }(x) \\ &=&\left( \sin x\right) ^{\ln x}\frac{\ln \left( \sin x\right) \sin x+\left( \left( \ln x\right) \cos x\right) x}{x\sin x}.\tag{9} \end{eqnarray*}$$

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$e^x$ and $\ln x$ are inverse functions. Therefore, $e^{\ln u}=u$. Then for the second part, $(a^b)^c=a^{bc}$. The function should now be in a form where you can take the derivative. –  Mike May 4 '12 at 15:40
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@Jordan No, $e^{\ln x}=x$ –  Américo Tavares May 4 '12 at 16:41
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@Jordan I've thought evaluating the derivative of $e^{\text{some function of } x}$. So I converted the given function $y=(\sin x)^{\ln x}$ to that form. I started by writing $\sin x$ as $e^{\ln(\sin x)}$. –  Américo Tavares May 4 '12 at 16:55
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@Jordan $\ln x$ is the exponent of $\sin x$. So, after converting $\sin x$ to $e^{\ln (\sin x)}$, we have to multiply the exponents $\ln x$ and $\ln (\sin x)$. –  Américo Tavares May 4 '12 at 17:13
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@Jordan Because of the power rule $(u^v)^w=u^{v\cdot w}$ mentioned in 2, applied to $u=e, v=\ln(\sin x), w=\ln x.$ –  Américo Tavares May 4 '12 at 17:39
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You can either use the definition that says that $$a^b = e^{b\ln(a)}$$ and use the chain rule with $$y(x) = (\sin x)^{\ln x} = e^{(\ln(x))(\ln(\sin x))}$$ or you can use logarithmic differentiation.

If $y = (\sin x)^{\ln x}$, then taking logarithms on both sides we get $$\ln y = \ln\left((\sin x)^{\ln x}\right) = (\ln x)\ln(\sin x).$$ Now using implicit differentiation we have: $$\begin{align*} \frac{d}{dx}\ln y &= \frac{d}{dx}\left( (\ln x)\ln(\sin x)\right)\\ \frac{1}{y}\frac{dy}{dx} &= \left(\ln x\right)'\ln(\sin x) + (\ln x)\left(\ln (\sin x)\right)'\\ \frac{y'}{y} &= \frac{1}{x}\ln(\sin x) + (\ln x)\left(\frac{1}{\sin x}(\sin x)'\right)\\ \frac{y'}{y} &= \frac{\ln\sin x}{x} + \frac{(\ln x)\cos x}{\sin x}\\ \frac{y'}{y} &= \frac{\ln \sin x}{x} + \ln x\cot x\\ y' &= y\left(\frac{\ln \sin x}{x} + \ln x\cot x\right)\\ y' &= (\sin x)^{\ln x}\left(\frac{\ln \sin x}{x} + \ln x\cot x\right). \end{align*}$$

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