Question

find the derivative of y= log e sin x​

Answer 1

Question :

Find derivate of

$\tt \: y = log_{e}(x) \times \sin \: x$

Solution :

Let $\sf\:u=\log_{e}\:x$

and v = sinx

then : y = u×v

Now Differentiate with respect to x ,by chain rule

$\tt \dfrac{dy}{dx} = \sin \: x \times \dfrac{ d\log_{e}(x)}{dx} + \log_{e}(x) \times \dfrac{ d\sin \: x}{dx}$

$\implies \tt \dfrac{dy}{dx} = \sin \: x \times \dfrac{1}{x} + log_{e}(x) \times \cos \: x$

$\implies \tt \dfrac{dy}{dx} = \dfrac{ \sin \:x}{x} + \log_{e}(x) \times \cos \: x$

It is the required solution !

Answer 2

Given :

Function

$\rm\:y=\log_{e}(x)\times\sin\:x$

${\red{\boxed{\large{\bold{Product\:Rule}}}}}$

Let u = f(x) and v = g(x) ,then

$\sf\:\dfrac{d(uv)}{dx} =u\dfrac{dv}{dx}+v\dfrac{du}{dx}$

Solution :

$\sf\:y=\log_{e}(x)\times\sin\:x$

Now Differentiate with respect to x ,by chain rule

$\sf\implies\dfrac{dy}{dx} = \sin \: x \times \dfrac{d\log_{e}(x)}{dx}+\log_{e}(x) \times \dfrac{ d\sin \: x}{dx}$

$\implies\sf\dfrac{dy}{dx} = \sin \: x \times \dfrac{1}{x} + log_{e}(x) \times \cos \: x$

$\implies\bf\dfrac{dy}{dx}=\dfrac{\sin\:x}{x}+\log_{e}(x)\times\cos\:x$

$\rule{200}2$

${\underline{\sf{Formula's}}}$

$1)\sf\:\frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}$

$2)\sf\:\frac{d(constant)}{dx} = 0$

$3) \sf\dfrac{d( \tan \: x)}{dx} = \sec{}^{2}\: x$

$\sf4) \dfrac{d(e {}^{x}) }{dx} = {e}^{x}$