The value of log _{e} 2 cdot frac{d}{dx}(log | vedclass

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

Question

(D) Let $y = \log _{\cos x} \operatorname{cosec} x$. Using the change of base formula for logarithms,$y = \frac{\ln(\operatorname{cosec} x)}{\ln(\cos

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Let $y = \log _{\cos x} \operatorname{cosec} x$. Using the change of base formula for logarithms,$y = \frac{\ln(\operatorname{cosec} x)}{\ln(\cos x)} = \frac{-\ln(\sin x)}{\ln(\cos x)}$. Differentiating with respect to $x$ using the quotient rule: $\frac{dy}{dx} = -\left[ \frac{\ln(\cos x) \cdot \frac{d}{dx}(\ln(\sin x)) - \ln(\sin x) \cdot \frac{d}{dx}(\ln(\cos x))}{(\ln(\cos x))^2} ight]$. $\frac{dy}{dx} = -\left[ \frac{\ln(\cos x) \cdot \cot x - \ln(\sin x) \cdot (-	an x)}{(\ln(\cos x))^2} ight]$. At $x = \frac{\pi}{4}$,$\sin x = \cos x = \frac{1}{\sqrt{2}}$,so $\ln(\sin x) = \ln(\cos x) = \ln(2^{-1/2}) = -\frac{1}{2} \ln 2$. Substituting these values: $\left. \frac{dy}{dx} ight|_{x=\frac{\pi}{4}} = -\left[ \frac{(-\frac{1}{2} \ln 2)(1) - (-\frac{1}{2} \ln 2)(-1)}{(-\frac{1}{2} \ln 2)^2} ight] = -\left[ \frac{-\frac{1}{2} \ln 2 - \frac{1}{2} \ln 2}{\frac{1}{4} (\ln 2)^2} ight] = -\left[ \frac{-\ln 2}{\frac{1}{4} (\ln 2)^2} ight] = \frac{4}{\ln 2}$. Finally,the value of $\log _{e} 2 \cdot \frac{dy}{dx}$ at $x = \frac{\pi}{4}$ is $\ln 2 \cdot \frac{4}{\ln 2} = 4$.

The value of $\log _{e} 2 \cdot \frac{d}{dx}(\log _{\cos x} \operatorname{cosec} x)$ at $x=\frac{\pi}{4}$ is.

Similar Questions

The differential coefficient of the function $\log_e \left( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right)$ with respect to $x$ is:

If $y=2^{\log x}$,then $\frac{d y}{d x}$ is

If $y = \log \left[a^{3x} \left(\frac{5-x}{x+4}\right)^{\frac{3}{4}}\right]$,then $\frac{dy}{dx} = $

If the angle between the curves $y = a^x$ and $y = b^x$ is $\alpha$,find the modulus of $\tan \alpha$.

The derivative of $(\log x)^{\sin x}$ with respect to $\cos x$ at $x=\frac{\pi}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module