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Hey guys I'm trying to find the derivative of this equation using logarithmic differentiation but I'm having some trouble. Wolfram Alpha is giving me different answers and I'm having difficulty understanding the steps it spits out. Could somebody take a look at what I'm doing and tell me where I went wrong?

The problem I need to differentiate is: $$y = (\sin x)^{\ln x}$$

My first step is to take the logarithm of both sides and factor out the exponent. $$\ln(y) = \ln x * \ln(\sin x)$$

Next I try to differentiate both sides, using the rules for logarithms: $$\frac{1}{y} * y' = \frac{1}{x}\frac{1}{\sin x}\cos x \Rightarrow \frac{1}{y}*y' = \frac{\cos x}{x \sin x}$$

Which finally renders: $$\frac{dy}{dx} = \frac{\cos x}{x\sin x}(\sin x)^{\ln x}$$

But this doesn't seem to be the right answer, could somebody help me out?

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  • $\begingroup$ Use the product rule to differentiate $\ln x\cdot \ln \sin x$ $\endgroup$ – paw88789 Jul 17 '14 at 15:46
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You need to use the product rule when differentiating:

$$\frac{d}{dx}(fg)=\frac{df}{dx}g+f\frac{dg}{dx}$$

So if $\ln{y}=\ln{x}\cdot\ln{(\sin{x})}$, for the purposes of differentiating the right hand side we have that $f=\ln{x}$ and that $g=\ln{(\sin{x})}$. So

$$\frac{y'}{y}=\frac{d}{dx}(\ln{x})\cdot\ln{(\sin{x})}+\ln{x}\cdot\frac{d}{dx}(\ln{(\sin{x})})$$

Thus

$$\frac{y'}{y}=\frac{1}{x}\cdot\ln{(\sin{x})}+\ln{x}\cdot\frac{\cos{x}}{\sin{x}}$$

Simplifying

$$y'=y\big(\frac{\ln{(\sin{x})}}{x}+\ln{x}\cot{x} \big)=(\sin{x})^{\ln{x}}\big(\frac{\ln{(\sin{x})}}{x}+\ln{x}\cot{x} \big)$$

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You forgot to use the product rule, $$\frac{d}{dx}\left(\ln x\cdot\ln(\sin x)\right)=\ln x\cdot\frac{d}{dx}\ln(\sin x)+\left(\frac{d}{dx}\ln x\right)\cdot\ln(\sin x)\ne\frac{1}{x}\frac{1}{\sin x}\cos x$$

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You're differentiating a product, which is $\ln(x) \ln(\sin(x))$. So it's $(uv)' = u'v + uv'$

Here: $\frac{\ln(\sin(x))}{x} + \frac{\ln x \cos(x)}{\sin(x)}$

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