# The derivative of $f\left( \tan x \right)$ with respect to $g\left( \sec x \right)$ at $x=\dfrac{\pi }{4}$, where $f'\left( 1 \right)=2;g'\left( \sqrt{2} \right)=4$ is:

(a) $\dfrac{1}{\sqrt{2}}$

(b) $\sqrt{2}$

(c) 1

(d) 0

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**Hint:**We have to find the derivative of $f\left( \tan x \right)$ with respect to $g\left( \sec x \right)$ at $x=\dfrac{\pi }{4}$. For that first of all differentiate $f\left( \tan x \right)$ with respect to x then differentiate $g\left( \sec x \right)$ with respect to x and then divide both the results of the differentiation. And after that put $x=\dfrac{\pi }{4}$ in place of x which will give you the required answer.

**Complete step-by-step solution:**We are asked to differentiate $f\left( \tan x \right)$ with respect to $g\left( \sec x \right)$ at $x=\dfrac{\pi }{4}$. For that, first of all we are going to derivate $f\left( \tan x \right)$ with respect to x and then differentiate $g\left( \sec x \right)$ with respect to x.

Differentiating $f\left( \tan x \right)$ with respect to x we get,

$\dfrac{df\left( \tan x \right)}{dx}=f'\left( \tan x \right){{\sec }^{2}}x$………. Eq. (1)

We have differentiated the above using chain rule. First of all we have differentiated $f\left( \tan x \right)$ then has multiplied the result of differentiation of $\tan x$ with respect to x.

Differentiating $g\left( \sec x \right)$ with respect to x we get,

$\dfrac{dg\left( \sec x \right)}{dx}=g'\left( \sec x \right)\left( \sec x\tan x \right)$…………. Eq. (2)

The above differentiation is also done using chain rule in the same way as we have shown for $f\left( \tan x \right)$.

Now, dividing eq. (1) by eq. (2) we get,

$\begin{align}

& \dfrac{\dfrac{df\left( \tan x \right)}{dx}}{\dfrac{dg\left( \sec x \right)}{dx}}=\dfrac{f'\left( \tan x \right){{\sec }^{2}}x}{g'\left( \sec x \right)\left( \sec x\tan x \right)} \\

& \Rightarrow \dfrac{df\left( \tan x \right)}{dg\left( \sec x \right)}=\dfrac{f'\left( \tan x \right){{\sec }^{2}}x}{g'\left( \sec x \right)\left( \sec x\tan x \right)} \\

\end{align}$

Now, we have to find the derivative of $f\left( \tan x \right)$ with respect to $g\left( \sec x \right)$ at $x=\dfrac{\pi }{4}$ so substituting $x=\dfrac{\pi }{4}$ in the above equation we get,

$\dfrac{df\left( \tan \dfrac{\pi }{4} \right)}{dg\left( \sec \dfrac{\pi }{4} \right)}=\dfrac{f'\left( \tan \dfrac{\pi }{4} \right){{\sec }^{2}}\left( \dfrac{\pi }{4} \right)}{g'\left( \sec \dfrac{\pi }{4} \right)\left( \sec \dfrac{\pi }{4}\tan \dfrac{\pi }{4} \right)}$

We know that,

$\begin{align}

& \tan \dfrac{\pi }{4}=1; \\

& \sec \dfrac{\pi }{4}=\sqrt{2} \\

\end{align}$

Substituting these values in the differentiation equation we get,

$\dfrac{df\left( 1 \right)}{dg\left( \sqrt{2} \right)}=\dfrac{f'\left( 1 \right){{\left( \sqrt{2} \right)}^{2}}}{g'\left( \sqrt{2} \right)\left( \sqrt{2} \right)}$………… Eq. (3)

It is also given in the above problem that:

$f'\left( 1 \right)=2;g'\left( \sqrt{2} \right)=4$

Substituting the above values in eq. (3) we get,

$\begin{align}

& \dfrac{df\left( 1 \right)}{dg\left( \sqrt{2} \right)}=\dfrac{2{{\left( \sqrt{2} \right)}^{2}}}{4\left( \sqrt{2} \right)} \\

& \Rightarrow \dfrac{df\left( 1 \right)}{dg\left( \sqrt{2} \right)}=\dfrac{2\left( 2 \right)}{4\sqrt{2}} \\

\end{align}$

4 will be cancelled from the numerator and denominator and we are left with:

$\dfrac{df\left( 1 \right)}{dg\left( \sqrt{2} \right)}=\dfrac{1}{\sqrt{2}}$

Hence, the differentiation of $f\left( \tan x \right)$ with respect to $g\left( \sec x \right)$ at $x=\dfrac{\pi }{4}$ is $\dfrac{1}{\sqrt{2}}$.

**Hence, the correct option is (a).**

**Note:**You might have thought how we know that we should separately differentiate $f\left( \tan x \right)\And g\left( \sec x \right)$ with respect to x and then divide each other because if you put $x=\dfrac{\pi }{4}$ in $f\left( \tan x \right)\And g\left( \sec x \right)$ you will get $f\left( 1 \right)\And g\left( \sqrt{2} \right)$ and if we differentiate both of these we will get values $f'\left( 1 \right)\And g'\left( \sqrt{2} \right)$ and values of $f'\left( 1 \right)\And g'\left( \sqrt{2} \right)$ is already given in the problem.

And the only way to differentiate the following is if we divide dx in numerator and denominator:

$\dfrac{df\left( \tan x \right)}{dg\left( \sec x \right)}$

Dividing dx in numerator and denominator we get,

$\dfrac{\dfrac{df\left( \tan x \right)}{dx}}{\dfrac{dg\left( \sec x \right)}{dx}}$

And the above derivative means we have to separately differentiate $f\left( \tan x \right)\And g\left( \sec x \right)$.