If y = frac{e^{2x} cos x}{x sin x},then frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $y = \frac{e^{2x} \cos x}{x \sin x} = \frac{e^{2x} \cot x}{x}$.Taking the natural logarithm on both sides:$\ln y = \ln(e^{2x}) + \ln(\cot x)

Vedclass · Accepted Answer

$\frac{e^{2x}[(2x - 1)\cot x - x \csc^2 x]}{x^2}$

Vedclass · Answer

(A) Given $y = \frac{e^{2x} \cos x}{x \sin x} = \frac{e^{2x} \cot x}{x}$.Taking the natural logarithm on both sides:$\ln y = \ln(e^{2x}) + \ln(\cot x) - \ln(x) = 2x + \ln(\cot x) - \ln x$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = 2 + \frac{1}{\cot x} \cdot (-\csc^2 x) - \frac{1}{x}$.$\frac{1}{y} \frac{dy}{dx} = 2 - 	an x \cdot \csc^2 x - \frac{1}{x} = 2 - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} - \frac{1}{x} = 2 - \frac{1}{\sin x \cos x} - \frac{1}{x} = 2 - 2 \csc(2x) - \frac{1}{x}$.Alternatively,using the quotient rule on $y = \frac{e^{2x} \cot x}{x}$:$\frac{dy}{dx} = \frac{x \cdot \frac{d}{dx}(e^{2x} \cot x) - e^{2x} \cot x \cdot \frac{d}{dx}(x)}{x^2}$.$\frac{dy}{dx} = \frac{x [2e^{2x} \cot x + e^{2x} (-\csc^2 x)] - e^{2x} \cot x}{x^2}$.$\frac{dy}{dx} = \frac{e^{2x} [2x \cot x - x \csc^2 x - \cot x]}{x^2}$.$\frac{dy}{dx} = \frac{e^{2x} [(2x - 1) \cot x - x \csc^2 x]}{x^2}$.

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

Similar Questions

$y=\frac{\sqrt[3]{1+3 x} \sqrt[4]{1+4 x} \sqrt[5]{1+5 x}}{\sqrt[7]{1+7 x} \sqrt[8]{1+8 x}}$. Then $\frac{d y}{d x}$ at $x=0$ is

If $\frac{d}{d x}\left(\frac{x \cdot 2^x-x}{1-\cos x}\right)=\left(\frac{x \cdot 2^x-x}{1-\cos x}\right)(f(x)+\log 2)$,then $f(x)=$

If $y=\frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x}$,then $\frac{d y}{d x}=$

If $f(x) = (|x|)^{|\sin x|}$,then $f'\left( -\frac{\pi}{4} \right) = $

Differentiate the function with respect to $x$: $(\sin x)^{x} + \sin^{-1} \sqrt{x}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module