Question

if y=log (x sin 2x),find dy/dx​

Answer 1

       $\underline{\bold{Solution:}}$

       $y = log(xsin2x)$

$\implies \dfrac{dy}{dx} = \dfrac{d}{dx}[log(xsin2x)]$

       $\text{Applying chain rule which states that:}$

       $\dfrac{d}{dy}[f(g(x))]= f'(g(x))\times g'(x)$

$\implies \dfrac{dy}{dx} = \dfrac{d}{dx}[log(xsin2x)]\times\dfrac{d}{dx}[xsin2x]$

$\implies \dfrac{dy}{dx} = \dfrac{1}{xsin2x}\times\dfrac{d}{dx}[xsin2x]$

       $\text{Applying the product rule which states that:}$

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

$\implies \dfrac{dy}{dx} = \dfrac{1}{xsin2x}\times[x\dfrac{d(sin2x)}{dx} + sin2x(\dfrac{dx}{dx})]$

$\implies \dfrac{dy}{dx} = \dfrac{1}{xsin2x}\times[x\dfrac{d(sin2x)}{dx} + sin2x]$

       $\text{Applying chain rule}$

$\implies \dfrac{dy}{dx} = \dfrac{1}{xsin2x}\times[x\{\dfrac{d(sin2x)}{dx}\times\dfrac{d(2x)}{dx}\} + sin2x]$

       $\text{Applying constant rule which states that:}$

       $\dfrac{d[kf(x)]}{dx} = k\dfrac{d[f(x)]}{dx}$

$\implies \dfrac{dy}{dx} = \dfrac{1}{xsin2x}\times[x(cos2x\times2) + sin2x]$

$\implies \dfrac{dy}{dx} = \dfrac{1}{xsin2x}\times[2xcos2x + sin2x]$

$\implies \dfrac{dy}{dx} = \dfrac{2xcos2x + sin2x}{xsin2x}$

$\implies \dfrac{dy}{dx} = \dfrac{2xcos2x}{xsin2x} + \dfrac{sin2x}{xsin2x}$

$\implies \dfrac{dy}{dx} = 2cot2x + \dfrac{1}{x}$

$\implies \boxed{\dfrac{dy}{dx} = \dfrac{2xcotx+1}{x}}$