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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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    @Jordan $y=(\sin x)^{\ln x}$ means that $y$ is $\sin x$ raised to the $\ln x$ power, while $\sin x^{\ln x}=\sin (x^{\ln x})$.2012-05-04

2 Answers 2

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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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    @Jordan In the meantime I rewrote it. You you need I'll roll it back.2012-05-04
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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: $\begin{align*} \frac{d}{dx}\ln y &= \frac{d}{dx}\left( (\ln x)\ln(\sin x)\right)\\ \frac{1}{y}\frac{dy}{dx} &= \left(\ln x\right)'\ln(\sin x) + (\ln x)\left(\ln (\sin x)\right)'\\ \frac{y'}{y} &= \frac{1}{x}\ln(\sin x) + (\ln x)\left(\frac{1}{\sin x}(\sin x)'\right)\\ \frac{y'}{y} &= \frac{\ln\sin x}{x} + \frac{(\ln x)\cos x}{\sin x}\\ \frac{y'}{y} &= \frac{\ln \sin x}{x} + \ln x\cot x\\ y' &= y\left(\frac{\ln \sin x}{x} + \ln x\cot x\right)\\ y' &= (\sin x)^{\ln x}\left(\frac{\ln \sin x}{x} + \ln x\cot x\right). \end{align*}$