Question
Diffrentiate the following w.r.t.x log(sec 3x+ tan 3x)
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}[\log (\sec 3 x+\tan 3 x)] \\ & =\frac{1}{\sec 3 x+\tan 3 x} \cdot \frac{d}{d x}(\sec 3 x+\tan 3 x) \\ & =\frac{1}{\sec 3 x+\tan 3 x} \times\left[\frac{d}{d x}(\sec 3 x)+\frac{d}{d x}(\tan 3 x)\right] \\ & =\frac{1}{\sec 3 x+\tan 3 x} \times\left[\sec 3 x \tan 3 x \cdot \frac{d}{d x}(3 x)+\sec ^2 3 x \cdot \frac{d}{d x}(3 x)\right]\end{aligned}$
$=\frac{1}{\sec 3 x+\tan 3 x} \times\left[\sec 3 x \tan 3 x \times 3+\sec ^2 3 x \times 3\right]$
$=\frac{3 \sec 3 x(\tan 3 x+\sec 3 x)}{\sec 3 x+\tan 3 x}=3 \sec 3 x$
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