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

Differentiate $\sqrt{\frac{(x-3)(x^{2}+4)}{3x^{2}+4x+5}}$ with respect to $x$.

If $y = \sin x \cdot \sin 2x \cdot \sin 3x \cdot \ldots \cdot \sin nx$,then $y^{\prime}$ is

If $y = (1 + x)^x$,then $\frac{dy}{dx} = $

Differentiate the function with respect to $x$: $(\log x)^{x}+x^{\log x}$

If $y = (\tan x)^{\cot x}$,then $\frac{dy}{dx} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module