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

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(D) ધારો કે $y = \log \sqrt{\frac{1+\sin x}{1-\sin x}}$. લઘુગણકના ગુણધર્મોનો ઉપયોગ કરતા,$y = \frac{1}{2} \log \left( \frac{1+\sin x}{1-\sin x} \right)

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(D) ધારો કે $y = \log \sqrt{\frac{1+\sin x}{1-\sin x}}$. લઘુગણકના ગુણધર્મોનો ઉપયોગ કરતા,$y = \frac{1}{2} \log \left( \frac{1+\sin x}{1-\sin x} \right) = \frac{1}{2} [\log(1+\sin x) - \log(1-\sin x)]$. $x$ ની સાપેક્ષમાં વિકલન કરતા: $\frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{1+\sin x} \cdot \cos x - \frac{1}{1-\sin x} \cdot (-\cos x) \right]$ $\frac{dy}{dx} = \frac{1}{2} \left[ \frac{\cos x}{1+\sin x} + \frac{\cos x}{1-\sin x} \right]$ $\frac{dy}{dx} = \frac{1}{2} \cos x \left[ \frac{1-\sin x + 1+\sin x}{(1+\sin x)(1-\sin x)} \right]$ $\frac{dy}{dx} = \frac{1}{2} \cos x \left[ \frac{2}{1-\sin^2 x} \right] = \frac{\cos x}{\cos^2 x} = \sec x$.

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

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$\frac{d}{d x}\left(\log \left(\frac{1}{x}\right)+\log \left(\frac{1}{x^2}\right)+\log\left(\frac{1}{x^3}\right)\right) = \text{ . . . . . . }$,$x > 1$

વિકલન શોધો: $\frac{d}{dx}[(\log_e x)(\log_a x)]$

$\frac{d}{dx} \left[ \log \left\{ e^x \left( \frac{x + 2}{x - 2} \right)^{3/4} \right\} \right]$ ની કિંમત શોધો.

$y = \log \left( \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+x} \right) \Rightarrow \frac{dy}{dx} = $

$\frac{d}{dx} \left\{ \log \left( \frac{e^x}{1 + e^x} \right) \right\} = $

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