Question

What is the derivative of $\arctan \left(2x\right)$ ?

Answer 1

Hint: In order to find the value of $\arctan \left(2x\right)$ , we should know what $\arctan$ is. $\arctan$ stands for arctangent. Arctangent is nothing but the inverse of the tangent function of $x$ , when $x$ is real. For example, when the tangent of $y$ is equal to $x$ that is $\tan y=x$ . Then $\arctan x$ is equal to the inverse tangent function, that is $\arctan x={\tan }^{-1}x=y$ .

Complete step-by-step answer:
We are given with the value $\arctan \left(2x\right)$ , let it be $y$ that gives $y=\arctan \left(2x\right)$ .
Considering the value of $2x$ to be $u$ , which gives:
 $2x=u$
Differentiating the above equation with respect to $u$ and we get:
 $2{\displaystyle \frac{dx}{du}}={\displaystyle \frac{du}{du}}$
Since, we know that ${\displaystyle \frac{du}{du}}=1$ , so:
 $2{\displaystyle \frac{dx}{du}}=1$
Dividing both the sides by $2$ :
 ${\displaystyle \frac{2{\displaystyle \frac{dx}{du}}}{2}}={\displaystyle \frac{1}{2}}$
 $\Rightarrow {\displaystyle \frac{dx}{du}}={\displaystyle \frac{1}{2}}$ ……(1)
Substituting $u=2x$ in the above equation $y=\arctan \left(2x\right)$ , we get:
 $y=\arctan \left(u\right)$
Since, we know that $\arctan$ is nothing but the inverse of the tangent function, so the $\arctan$ can be written as $ta{n}^{-1}$ .
So, from this we are writing our equation as:
 $y=ta{n}^{-1}\left(u\right)$
Now, for the derivative of the function, derivating both the sides of the equation with respect to $u$ , we get:
 ${\displaystyle \frac{dy}{du}}={\displaystyle \frac{d\left({\tan }^{-1}u\right)}{du}}$
From the trigonometric formulas for derivation, we know that:
 ${\displaystyle \frac{d\left({\tan }^{-1}x\right)}{dx}}={\displaystyle \frac{1}{1+x^2}}$
So, from this we get:
 ${\displaystyle \frac{dy}{du}}={\displaystyle \frac{d\left({\tan }^{-1}u\right)}{du}}$
 $\Rightarrow {\displaystyle \frac{dy}{du}}={\displaystyle \frac{1}{1+u^2}}$ ………….(2)
Now, dividing equation (2) by (1), and we get:
 ${\displaystyle \frac{{\displaystyle \frac{dy}{du}}}{{\displaystyle \frac{dx}{du}}}}={\displaystyle \frac{{\displaystyle \frac{1}{1+u^2}}}{{\displaystyle \frac{1}{2}}}}$
Since, on the left side the denominators are same, so cancelling them out and multiplying $2$ to the numerator and denominator of right side:
 ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{{\displaystyle \frac{1}{1+u^2}}\times 2}{{\displaystyle \frac{1}{2}}\times 2}}$
 $\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{2}{1+u^2}}$
Substituting the value of $u$ that is $2x=u$ in the above equation, and we get:
 ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{2}{1+{\left(2x\right)}^2}}$
 $\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1}{1+4x^2}}$
Since, $y=\arctan \left(2x\right)$ , therefore ${\displaystyle \frac{d\left(\arctan \left(2x\right)\right)}{dx}}={\displaystyle \frac{1}{1+4x^2}}$ .
Hence, the derivative of $\arctan \left(2x\right)={\displaystyle \frac{1}{1+4x^2}}$ .

Note: The method used above to consider $2x=u$ , then differentiating them separately with respect to $u$ , then for ${\tan }^{-1}u$ is known as the chain rule. It can also be written as ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{dy}{du}}\times {\displaystyle \frac{du}{dx}}$ .
It’s important to remember the basic formula of trigonometry and trigonometric identities to solve these kinds of questions.