Question

please differentiate the above.​

Answer 1

Step-by-step explanation:

d/dx [ sin x (ln x) ]

consider y = sin(x) ln(x)

dy/dx = [d/dx(sin(x))] ln(x)+sin(x) d/dx(ln(x))

dy/dx = cos(x)ln(x)+sin(x)(1/x)

Answer 2

Solution:-

$\implies \: \frac{d}{dx} \sin\: x \: ln \: x$

We know that formula of

$\frac{d}{dx} fx.gx = gx\frac{d}{dx}fx + fx \: \frac{d}{dx}gx$

$\implies \: ln \: x \: \frac{d}{dx} \sin \: x + \sin \: x \: \frac{d}{dx} ln \: x$

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we know that

d/dx sinx=cos x

d/dx in x=1/x

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$\implies \: ln \: x \:. \cos \: x + \sin \: x . \frac{1}{x}$

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I hope it helps you