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y = sinx cosx logx. Find dy/dx.

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\(y = sin x \;cosx\;logx\)

\(\therefore \frac{dy}{dx} = sin x \frac d{dx} (cos x \; log x) + (\frac d{dx} sin x) (cos x\; log x)\)

\(= sin x \left(cos \, x \frac d{dx}log x + \left(\frac d{dx}cos x\right)log x\right) + cos x(cosx\; logx)\)

\(=\frac{sinx\;cosx}x - sin^2 x\; log x + cos^2 x\; logx\)

\(=\frac{2sinx\;cos x}{2x} + (cos^2x-sin^2x)logx\)

\(= \frac{sin(2x)}{2x} + cos(2x) log x\)

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