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.