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# Find $\dfrac{dy}{dx}$ where $y=\log \left( \sec x \right)$ for $0\le x\le \dfrac{\pi }{2}$.  Verified
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Assume sec x as t and differentiate it, we get value of dt.By substituting value of sec x as t in given equation we get log t and differentiate this equation with respect to x by using product rule and substitute the values of t and dt to get required answer.

Given, $y=\log \left( \sec x \right)$.
$\Rightarrow t=\sec x$
\begin{align} & dt=\sec x\tan xdx \\ & \Rightarrow \dfrac{dt}{dx}=\sec x\tan x \\ & y=\log t \\ & \dfrac{dy}{dx}=\left( \dfrac{dy}{dt} \right)\times \left( \dfrac{dt}{dx} \right) \\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dt}\left( \log t \right)\times \dfrac{d}{dx}\left( \sec x \right) \\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{t}\times \sec x\tan x \\ \end{align}
\begin{align} & \dfrac{dy}{dx}=\dfrac{1}{\sec x}\times \sec x\tan x \\ & \therefore \dfrac{dy}{dx}=\tan x \\ \end{align}
$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dm}\times \dfrac{dm}{dx}$
Also, don’t get confused by the fact that it is mentioned $x\in \left[ 0,\dfrac{\pi }{2} \right]$.