frac{d}{dx}{ log (sec x + tan x)} = | vedclass

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(B) Let $y = \log (\sec x + 	an x)$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{\sec x + 	an x} \cdot \fr

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$\sec x$

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(B) Let $y = \log (\sec x + 	an x)$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{\sec x + 	an x} \cdot \frac{d}{dx}(\sec x + 	an x)$Since $\frac{d}{dx}(\sec x) = \sec x 	an x$ and $\frac{d}{dx}(	an x) = \sec^2 x$,we have:$\frac{dy}{dx} = \frac{\sec x 	an x + \sec^2 x}{\sec x + 	an x}$Factoring out $\sec x$ from the numerator:$\frac{dy}{dx} = \frac{\sec x (	an x + \sec x)}{\sec x + 	an x}$Canceling the common term $(\sec x + 	an x)$:$\frac{dy}{dx} = \sec x$.

$\frac{d}{dx}\{ \log (\sec x + \tan x)\} = $

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