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# How do you differentiate $\sec \left( {\arctan (x)} \right)$?

Last updated date: 11th Jun 2024
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Hint:Derivatives are defined as the varying rate of a function with respect to an independent variable. We cannot differentiate this directly. First we need to find the value of $\sec \left( {\arctan (x)} \right)$. After that we differentiate the obtained answer with respect to ‘x’. we know that $\tan \theta = \dfrac{{{\text{opposite side}}}}{{{\text{adjacent side}}}}$, $\sec \theta = \dfrac{{{\text{hypotenuse side}}}}{{{\text{adjacent side}}}}$ and using Pythagoras identity we can find the value of $\sec \left( {\arctan (x)} \right)$.

Complete step by step solution:
Given, $\sec \left( {\arctan (x)} \right)$
Let’s put $\theta = \arctan (x)$
Then we have $\sec \left( \theta \right)$
Now we took $\theta = \arctan (x)$,
Then we have $\tan \theta = x$
This can be rewrite as
$\tan \theta = \dfrac{x}{1}$
We know that $\tan \theta = \dfrac{{{\text{opposite side}}}}{{{\text{adjacent side}}}}$.
Let’s write a right angle triangle and we need to find hypotenuse side

We need hypotenuse, that is AC.
By Pythagoras identity we have
$\begin{gathered} A{C^2} = A{B^2} + B{C^2} \\ A{C^2} = {x^2} + 1 \\ AC = \sqrt {{x^2} + 1} \\ \end{gathered}$
Thus we have a hypotenuse side.
We know that $\sec \theta = \dfrac{{{\text{hypotenuse side}}}}{{{\text{adjacent side}}}}$
$\sec \theta = \dfrac{{\sqrt {{x^2} + 1} }}{1}$
That is we have,
$\sec \left( {\arctan (x)} \right) = \sqrt {{x^2} + 1}$
Now differentiating with respect to ‘x’
$\dfrac{d}{{dx}}\sec \left( {\arctan (x)} \right) = \dfrac{d}{{dx}}\sqrt {{x^2} + 1}$
$= \dfrac{d}{{dx}}\sqrt {{x^2} + 1}$
We know that $\dfrac{d}{{dx}}(\sqrt x ) = \dfrac{1}{{2\sqrt x }}\dfrac{{dx}}{{dx}}$ and here we assume ${x^2} + 1$ as one term ‘x’. Then we have
$= \dfrac{1}{{2\sqrt {{x^2} + 1} }}\dfrac{d}{{dx}}\left( {{x^2} + 1} \right)$
$= \dfrac{{2x}}{{2\sqrt {{x^2} + 1} }}$
$= \dfrac{x}{{\sqrt {{x^2} + 1} }}$
Thus the differentiation of $\sec \left( {\arctan (x)} \right)$ is $\dfrac{x}{{\sqrt {{x^2} + 1} }}$.

Note: We know the differentiation of ${x^n}$ with respect to ‘x’ is $\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}$. We also have different rules in the differentiation. Those are
$\bullet$Linear combination rule: The linearity law is very important to emphasize its nature with alternate notation. Symbolically it is specified as $h'(x) = af'(x) + bg'(x)$
$\bullet$Product rule: When a derivative of a product of two function is to be found, then we use product rule that is $\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$.
$\bullet$Chain rule: To find the derivative of composition function or function of a function, we use chain rule. That is $fog'({x_0}) = [(f'og)({x_0})]g'({x_0})$.
We use these rules depending on the given problem.