Question

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

Answer 1

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

Note: We know the differentiation of $$x^n$$ with respect to ‘x’ is $${\displaystyle \frac{d(x^n)}{dx}}=n.x^{n-1}$$. 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 $$h^{\prime }(x)=a{f}^{\prime }(x)+b{g}^{\prime }(x)$$
$$\bullet$$Product rule: When a derivative of a product of two function is to be found, then we use product rule that is $${\displaystyle \frac{dy}{dx}}=u\times {\displaystyle \frac{dv}{dx}}+v\times {\displaystyle \frac{du}{dx}}$$.
$$\bullet$$Chain rule: To find the derivative of composition function or function of a function, we use chain rule. That is $$fo{g}^{\prime }(x_0)=[(f^{\prime }og)(x_0)]g^{\prime }(x_0)$$.
We use these rules depending on the given problem.