Question

What is the derivative of \[y = \arctan (4x)\] ?

Answer 1

Hint: Here, the given question has a trigonometric function. We have to find the derivative or differentiated term of the function. First consider the function \[y\], then differentiate \[y\] with respect to \[x\] by using a standard differentiation formula of trigonometric ratio and use chain rule for differentiation. And on further simplification we get the required differentiate value.
In the trigonometry we have standard differentiation formula
the differentiation of \[\dfrac{d}{{dx}}(\tan x) = {\sec ^2}x\]
The trigonometric identities we have used here is \[1 + {\tan ^2}y = {\sec ^2}y\]

Complete step by step solution:
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Consider the given function
 \[ \Rightarrow y = \arctan (4x)\] ---------- (1)
This can be written as
 \[ \Rightarrow y = ta{n^{ - 1}}(4x)\]
Take \[\tan \] on both sides we have
 \[ \Rightarrow \tan y = \tan \left( {{{\tan }^{ - 1}}(4x)} \right)\]
On simplifying we get
 \[ \Rightarrow \tan y = 4x\] ----------(2)
Differentiate function y with respect to x
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {\tan y} \right) = \dfrac{d}{{dx}}\left( {4x} \right)\] -------(3)
As we know the formula \[\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x\], then
Equation (2) becomes
 \[ \Rightarrow {\sec ^2}y.\dfrac{{dy}}{{dx}} = 4\] ---------- (4)
As we know the trigonometric identities which is given as \[1 + {\tan ^2}y = {\sec ^2}y\], by the equation (2) the trigonometric identity
 \[ \Rightarrow 1 + {(4x)^2} = {\sec ^2}y\]
On simplifying we get
 \[ \Rightarrow 1 + 16{x^2} = {\sec ^2}y\] ----------(5)
By using the equation (5) and the equation (4) is written as
 \[ \Rightarrow \left( {1 + 16{x^2}} \right).\dfrac{{dy}}{{dx}} = 4\]
Take \[1 + 16{x^2}\] to the RHS and it is written as
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{{1 + 16{x^2}}}\]
Hence, it’s a required differentiated value.
So, the correct answer is “ \[\dfrac{4}{{1 + 16{x^2}}}\] ”.

Note: The student must know about the differentiation formulas for the trigonometry ratios and these differentiation formulas are standard. If the function is a product of two terms and the both terms are the function of x the we use the product rule of differentiation to the function.