Question

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

Answer 1

Hint: To find the derivative of the $\arctan \left( 2x \right)$, we will simply differentiate the given expression with respect to x. since we have only one variable in the given question, so it's not hard to differentiate the above expression. In mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to a ratio of two side lengths. The above given functions are trigonometric functions. The angles of sine, cosine, and tangent are the primary classification of functions of trigonometry.

Complete step by step solution:
Here we have to differentiate the $\arctan \left( 2x \right)$.
Actually differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another variable. Differentiation means finding the derivative of a function with respect to x.
Now,
$\Rightarrow \dfrac{d\left( {{\tan }^{-1}}2x \right)}{dx}$
Since, we know the value of $\dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}=\dfrac{1}{1+{{x}^{2}}}$ .
Here, at the place of x we have 2x. To solve the above function we will use chain rule because in this question we have two functions. The first is ${{\tan }^{-1}}$ and 2x. The chain rule is:
$\Rightarrow \dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=f\grave{\ }\left( g\left( x \right) \right)\cdot g\grave{\ }\left( x \right)$
Now, the derivative of the given function is:
\[\dfrac{2}{1+4{{x}^{2}}}\]
\[\begin{align}
  & \Rightarrow \dfrac{d\left( ta{{n}^{1}}2x \right)}{dx}=\dfrac{1}{1+{{\left( 2x \right)}^{2}}}\dfrac{d\left( 2x \right)}{dx} \\
 & \Rightarrow \dfrac{d\left( ta{{n}^{1}}2x \right)}{dx}=\dfrac{1}{1+4{{x}^{2}}}\cdot 2 \\
 & \Rightarrow \dfrac{d\left( ta{{n}^{1}}2x \right)}{dx}=\dfrac{2}{1+4{{x}^{2}}} \\
\end{align}\]
Here the differentiation of $2x$ is $2$. Hence we get the derivative of the given $\arctan \left( 2x \right)$ is \[\dfrac{2}{1+4{{x}^{2}}}\].

Note: Finding the derivative of any trigonometric function is not hard but the important thing is that we should be the direct derivative of some trigonometric function and any variables so that we can easily calculate the value of the derivative of any function. And to find the derivative we should know the rules of differentiation like quotient rule, chain rule, product rule and addition and subtraction rule. By these rules we can easily calculate derivatives.