[Math] Derivative with trig functions and ln trickery

calculusderivatives

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.

Best Answer

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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