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:
$2\dfrac{{dx}}{{du}} = \dfrac{{du}}{{du}}$
Since, we know that $\dfrac{{du}}{{du}} = 1$ , so:
$2\dfrac{{dx}}{{du}} = 1$
Dividing both the sides by $2$ :
$\dfrac{{2\dfrac{{dx}}{{du}}}}{2} = \dfrac{1}{2}$
$\Rightarrow \dfrac{{dx}}{{du}} = \dfrac{1}{2}$ ……(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:
$\dfrac{{dy}}{{du}} = \dfrac{{d\left( {{{\tan }^{ - 1}}u} \right)}}{{du}}$
From the trigonometric formulas for derivation, we know that:
$\dfrac{{d\left( {{{\tan }^{ - 1}}x} \right)}}{{dx}} = \dfrac{1}{{1 + {x^2}}}$
So, from this we get:
$\dfrac{{dy}}{{du}} = \dfrac{{d\left( {{{\tan }^{ - 1}}u} \right)}}{{du}}$
$\Rightarrow \dfrac{{dy}}{{du}} = \dfrac{1}{{1 + {u^2}}}$ ………….(2)
Now, dividing equation (2) by (1), and we get:
$\dfrac{{\dfrac{{dy}}{{du}}}}{{\dfrac{{dx}}{{du}}}} = \dfrac{{\dfrac{1}{{1 + {u^2}}}}}{{\dfrac{1}{2}}}$
Since, on the left side the denominators are same, so cancelling them out and multiplying $2$ to the numerator and denominator of right side:
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{1}{{1 + {u^2}}} \times 2}}{{\dfrac{1}{2} \times 2}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{1 + {u^2}}}$
Substituting the value of $u$ that is $2x = u$ in the above equation, and we get:
$\dfrac{{dy}}{{dx}} = \dfrac{2}{{1 + {{\left( {2x} \right)}^2}}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + 4{x^2}}}$
Since, $y = \arctan \left( {2x} \right)$ , therefore $\dfrac{{d\left( {\arctan \left( {2x} \right)} \right)}}{{dx}} = \dfrac{1}{{1 + 4{x^2}}}$ .
Hence, the derivative of $\arctan \left( {2x} \right) = \dfrac{1}{{1 + 4{x^2}}}$ .

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 $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$ .
It’s important to remember the basic formula of trigonometry and trigonometric identities to solve these kinds of questions.