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Find $\dfrac{dy}{dx}$ where $y=\log \left( \sec x \right)$ for $0\le x\le \dfrac{\pi }{2}$.

Last updated date: 24th May 2024
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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.

“Complete step-by-step answer:”
Given, $y=\log \left( \sec x \right)$.
Let us assume sec x to be t.
$\Rightarrow t=\sec x$
Differentiating both sides, we get:
  & 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 \\
Putting the value of t = sec x in the above equation we get,
  & \dfrac{dy}{dx}=\dfrac{1}{\sec x}\times \sec x\tan x \\
 & \therefore \dfrac{dy}{dx}=\tan x \\
Therefore, the answer is tan x.

Note: In the given question, we have used the product rule which is:
$\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dm}\times \dfrac{dm}{dx}$
Also, don’t get confused by the fact that it is mentioned $x\in \left[ 0,\dfrac{\pi }{2} \right]$.
It is mentioned to define the domain of log.