Question

If $y=\frac{e^{2x}\cos x}{x\sin x}$, then $\frac{dy}{dx}$is equal to ?

• A. $\frac{e^{2x}\left[(2x-1)cotx–x\cos e{c}^{2}x\right]}{x^2}$
• B. $\frac{e^{2x}\left[(2x-1)cotx–\cos e{c}^{2}x\right]}{x^2}$
• C. $\frac{e^{2x}\left[(2x-1)cotx+\cos e{c}^{2}x\right]}{x^2}$
• D. none of these

Answer 1

The correct option is A

$\frac{e^{2x}\left[(2x-1)cotx–x\cos e{c}^{2}x\right]}{x^2}$

Explanation for the correct option:

Find the value of $\frac{dy}{dx}$ :

Given,

$y=\frac{e^{2x}\cos x}{x\sin x}$

Now, differentiate the given function with respect to $x$

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{x\sin x\frac{d}{dx}{e}^{2x}\cos x–e^{2x}\cos x\frac{d}{dx}x\sin x}{{(x\sin x)}^2}\left[\because \frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{du}{dx}}-u{\displaystyle \frac{dv}{dx}}}{v^2}\right]\\ & =& \frac{x\sin x(e^{2x}\times 2\times \cos x–e^{2x}\sin x)–e^{2x}\cos x(\sin x+x\cos x)}{{(x\sin x)}^2}\mathbf{}\left[\because \frac{d\left(e^x\right)}{dx}=e^x,\frac{d\left(sinx\right)}{dx}=cosx\right]\\ & =& \frac{x2\sin x\cos x{e}^{2x}–e^{2x}x\sin 2x–e^{2x}\cos x\sin x–e^{2x}x\cos 2x}{{(x\sin x)}^2}\\ & =& \frac{x\sin 2x{e}^{2x}–x{e}^{2x}(\sin 2x+\cos 2x)–e^{2x}\sin x\cos x}{{(x\sin x)}^2}\\ & =& \frac{x{e}^{2x}\sin 2x–x{e}^{2x}–e^{2x}\sin x\cos x}{{(x\sin x)}^2}\\ & =& \frac{2x{e}^{2x}cotx–x{e}^{2x}\cos e{c}^{2}x–e^{2x}cotx}{x^2}\\ & =& \frac{e^{2x}\left[2xcotx–x\cos e{c}^{2}x–cotx\right]}{x^2}\\ & =& \frac{e^{2x}\left[(2x-1)cotx–x\cos e{c}^{2}x\right]}{x^2}\end{array}$

Hence, the correct option is A.