If f(x)=frac{(x+1) sinh x}{e^{2 x} tan x} and | vedclass

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

Question

(C) Given $f(x) = \frac{(x+1) \sinh x}{e^{2x} 	an x} = (x+1) \sinh x e^{-2x} \cot x$.Taking the natural logarithm on both sides: $\ln|f(x)| = \ln|x+1

Vedclass · Accepted Answer

$-2(1+\operatorname{cosec} 2 x)$

Vedclass · Answer

(C) Given $f(x) = \frac{(x+1) \sinh x}{e^{2x} 	an x} = (x+1) \sinh x e^{-2x} \cot x$.Taking the natural logarithm on both sides: $\ln|f(x)| = \ln|x+1| + \ln|\sinh x| - 2x + \ln|\cot x|$.Differentiating with respect to $x$: $\frac{f'(x)}{f(x)} = \frac{d}{dx}(\ln|x+1|) + \frac{d}{dx}(\ln|\sinh x|) + \frac{d}{dx}(-2x) + \frac{d}{dx}(\ln|\cot x|)$.$\frac{f'(x)}{f(x)} = \frac{1}{x+1} + \frac{\cosh x}{\sinh x} - 2 + \frac{-\operatorname{cosec}^2 x}{\cot x}$.Since $\frac{\cosh x}{\sinh x} = \operatorname{coth} x$ and $\frac{\operatorname{cosec}^2 x}{\cot x} = \frac{1}{\sin^2 x} \cdot \frac{\sin x}{\cos x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x} = 2 \operatorname{cosec} 2x$.Thus, $\frac{f'(x)}{f(x)} = \frac{1}{x+1} + \operatorname{coth} x - 2 - 2 \operatorname{cosec} 2x$.Comparing this with the given expression $\frac{1}{x+1} + \operatorname{coth} x + g(x)$, we get $g(x) = -2 - 2 \operatorname{cosec} 2x = -2(1 + \operatorname{cosec} 2x)$.

If $f(x)=\frac{(x+1) \sinh x}{e^{2 x} \tan x}$ and $\frac{f^{\prime}(x)}{f(x)}=\frac{1}{x+1}+\operatorname{coth} x+g(x)$, then $g(x)=$

Similar Questions

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

If $y = e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$,then $\frac{dy}{dx} = $

If $f(x) = \frac{e^{-x} \sin x}{\log_e x}$ and $f'(x) = f(x) \cdot g(x)$, then $g'(e) =$

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

If $\frac{d}{d x} \left[ \frac{(x+1)^2 \sqrt{x-1}}{(x+4)^3 e^x} \right] = f(x) \left[ \frac{2}{x+1} + \frac{1}{2(x-1)} - \frac{3}{x+4} - 1 \right]$,then $f(5) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module