Question

Find the derivative of

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Answer 1

We need to find:-

$\longrightarrow\sf{\dfrac{d}{dx}\left[\dfrac{\log x+e^{-x}}{\sin(2x)}\right]=\,?}$

We have, quotient rule,

$\longrightarrow\sf{\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)=\dfrac{\dfrac{d}{dx}\,\left[f(x)\right]\cdot g(x)-f(x)\cdot\dfrac{d}{dx}\,\left[g(x)\right]}{[g(x)]^2}}$

Here,

$\longrightarrow\sf{f(x)=\log x+e^{-x}}$

$\longrightarrow\sf{\dfrac{d}{dx}\,[f(x)]=\dfrac{d}{dx}\left[\log x+e^{-x}\right]}$

$\longrightarrow\sf{\dfrac{d}{dx}\,[f(x)]=\dfrac{d}{dx}[\log x]+\dfrac{d}{dx}\left[e^{-x}\right]}$

By chain rule,

$\longrightarrow\sf{\dfrac{d}{dx}\,[f(x)]=\dfrac{1}{x}+\dfrac{d[e^{-x}]}{d[-x]}\cdot\dfrac{d[-x]}{dx}}$

$\longrightarrow\sf{\dfrac{d}{dx}\,[f(x)]=\dfrac{1}{x}-e^{-x}}$

And,

$\longrightarrow\sf{g(x)=\sin(2x)}$

$\longrightarrow\sf{\dfrac{d}{dx}[g(x)]=\dfrac{d}{dx}[\sin(2x)]}$

By chain rule,

$\longrightarrow\sf{\dfrac{d}{dx}[g(x)]=\dfrac{d[\sin(2x)]}{d[2x]}\cdot\dfrac{d[2x]}{dx}}$

$\longrightarrow\sf{\dfrac{d}{dx}[g(x)]=2\cos(2x)}$

Hence by quotient rule,

$\longrightarrow\sf{\dfrac{d}{dx}\left(\dfrac{\log x+e^{-x}}{\sin(2x)}\right)=\dfrac{\sin(2x)\left[\dfrac{1}{x}-e^{-x}\right]-2\cos(2x)\left[\log x+e^{-x}\right]}{\sin^2(2x)}}$

$\longrightarrow\underline{\underline{\sf{\dfrac{d}{dx}\left(\dfrac{\log x+e^{-x}}{\sin(2x)}\right)=\dfrac{\sin(2x)-x\,\sin(2x)\,e^{-x}-2x\cos(2x)\log x-2x\cos(2x)\,e^{-x}}{x\,\sin^2(2x)}}}}$