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