frac{d}{dx} left( log sqrt{frac{1+sin x}{1-si | vedclass

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

Question

(D) Let $y = \log \sqrt{\frac{1+\sin x}{1-\sin x}}$. Using logarithmic properties,$y = \frac{1}{2} \log \left( \frac{1+\sin x}{1-\sin x} \right) = \fr

Vedclass · Accepted Answer

$\sec x$

Vedclass · Answer

(D) Let $y = \log \sqrt{\frac{1+\sin x}{1-\sin x}}$. Using logarithmic properties,$y = \frac{1}{2} \log \left( \frac{1+\sin x}{1-\sin x} \right) = \frac{1}{2} [\log(1+\sin x) - \log(1-\sin x)]$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{1+\sin x} \cdot \cos x - \frac{1}{1-\sin x} \cdot (-\cos x) \right]$ $\frac{dy}{dx} = \frac{1}{2} \left[ \frac{\cos x}{1+\sin x} + \frac{\cos x}{1-\sin x} \right]$ $\frac{dy}{dx} = \frac{1}{2} \cos x \left[ \frac{1-\sin x + 1+\sin x}{(1+\sin x)(1-\sin x)} \right]$ $\frac{dy}{dx} = \frac{1}{2} \cos x \left[ \frac{2}{1-\sin^2 x} \right] = \frac{\cos x}{\cos^2 x} = \sec x$.

$\frac{d}{dx} \left( \log \sqrt{\frac{1+\sin x}{1-\sin x}} \right) = $

Similar Questions

If $y = \log_{10} x + \log_{x} 10 + \log_{x} x + \log_{10} 10$,then $\frac{dy}{dx}$ is equal to

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

$\frac{d}{d x}(\log _{|x|} e) =$ . . . . . .

If $u = \log(\sqrt{x-1} - \sqrt{x+1})$ and $v = \sqrt{x+1} + \sqrt{x-1}$,then $\frac{du}{dv} = \dots$.

If $y = \log \left( \frac{1 + \sqrt{x}}{1 - \sqrt{x}} \right)$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module