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

Answer

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324.6k+ views

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:

$\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}$

Putting the value of t = sec x in the above equation we get,

$\begin{align}

& \dfrac{dy}{dx}=\dfrac{1}{\sec x}\times \sec x\tan x \\

& \therefore \dfrac{dy}{dx}=\tan x \\

\end{align}$

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.

“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:

$\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}$

Putting the value of t = sec x in the above equation we get,

$\begin{align}

& \dfrac{dy}{dx}=\dfrac{1}{\sec x}\times \sec x\tan x \\

& \therefore \dfrac{dy}{dx}=\tan x \\

\end{align}$

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.

Last updated date: 29th May 2023

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