Question

How do you take the derivative of \[{\tan ^2}\left( {4x} \right)\] ?

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 cos x is -sin x that is \[\dfrac{d}{{dx}}(\cos x) = - \sin x\]


Complete step-by-step answer:
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
  \[ \Rightarrow y = {\tan ^2}\left( {4x} \right)\] ---------- (1)
Differentiate function y with respect to x
 \[ \Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {{{\tan }^2}\left( {4x} \right)} \right)\] -------(2)
Here, we have to use the chain rule method to differentiate the above function.
As we know the formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\] , then
Equation (2) becomes
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = 2\tan \left( {4x} \right)\dfrac{d}{{dx}}\left( {\tan \left( {4x} \right)} \right)\] ---------- (3)
Using the standard differentiated formula of trigonometric ratio cosine is \[\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x\] , then equation (3) becomes
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = 2\tan \left( {4x} \right){\sec ^2}\left( {4x} \right)\dfrac{d}{{dx}}\left( {4x} \right)\]
Where, 4 is a constant and take it outside from the differentiated term, then
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = 2\tan \left( {4x} \right) \cdot {\sec ^2}\left( {4x} \right) \cdot 4\dfrac{d}{{dx}}\left( x \right)\] ------------(4)
As we know, the another standard formula: \[\dfrac{{dx}}{{dx}} = 1\]
Then equation (4) becomes
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = 2\tan \left( {4x} \right) \cdot {\sec ^2}\left( {4x} \right) \cdot 4\left( 1 \right)\]
On simplification, we get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = 8\tan \left( {4x} \right) \cdot {\sec ^2}\left( {4x} \right)\]
Hence, it’s a required differentiated value.
So, the correct answer is “ \[ \dfrac{{dy}}{{dx}} = 8\tan \left( {4x} \right) \cdot {\sec ^2}\left( {4x} \right)\] ”.

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 then we use the product rule of differentiation to the function.