Derivative of log (sec theta+tan theta) with | vedclass

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(B) Let $y_1 = \log (\sec 	heta + 	an 	heta)$.Then,$\frac{dy_1}{d	heta} = \frac{1}{\sec 	heta + 	an 	heta} \cdot (\sec 	heta 	an 	heta + \se

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(B) Let $y_1 = \log (\sec 	heta + 	an 	heta)$.Then,$\frac{dy_1}{d	heta} = \frac{1}{\sec 	heta + 	an 	heta} \cdot (\sec 	heta 	an 	heta + \sec^2 	heta)$.$\frac{dy_1}{d	heta} = \frac{\sec 	heta (	an 	heta + \sec 	heta)}{\sec 	heta + 	an 	heta} = \sec 	heta$.Let $y_2 = \sec 	heta$.Then,$\frac{dy_2}{d	heta} = \sec 	heta 	an 	heta$.We need to find $\frac{dy_1}{dy_2} = \frac{dy_1/d	heta}{dy_2/d	heta} = \frac{\sec 	heta}{\sec 	heta 	an 	heta} = \frac{1}{	an 	heta} = \cot 	heta$.At $	heta = \frac{\pi}{4}$,$\frac{dy_1}{dy_2} = \cot \frac{\pi}{4} = 1$.

Derivative of $\log (\sec \theta+\tan \theta)$ with respect to $\sec \theta$ at $\theta=\frac{\pi}{4}$ is

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