Question

What is the third derivative of \[\tan x\]?

Answer 1

Hint: In this problem we have to find the third derivative of the given trigonometric function, \[\tan x\]. Here the trigonometric function \[\tan x\] should be differentiated three times one by one. Then we can get the solution, the third derivative of \[\tan x\]. While differentiating the trigonometric function \[\tan x\]we must use some differentiation conditions and formulas.

Complete step by step solution:
Here we have to find the third derivative of \[\tan x\].
To get the third derivative of the trigonometric function \[\tan x\]we must differentiate it thrice and simplify it. Let us consider the trigonometric function \[\tan x\]as

\[\Rightarrow y=\tan x\]
Now differentiating it for the first time we get,
\[\Rightarrow y=\tan x\dfrac{dy}{dx}\]
By using the differentiation formula, differentiating \[\tan x\]we get,
\[\Rightarrow y'={{\sec }^{2}}x\]
We know that the first derivation of the trigonometric function \[\tan x\] is over. Now differentiating the first derivative we get second derivative, so
\[\Rightarrow y'={{\sec }^{2}}x\dfrac{d{{y}^{'}}}{dx}\]
Here we can use the formula,\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\] and then differentiating rest of the \[\sec x\] we get,
\[\Rightarrow y''=2\sec x\sec x\tan x\] \[\left[ \because \dfrac{d}{dx}\sec x=\sec x\tan x \right]\]
Now multiplying the similar terms we get the second derivation,
\[\Rightarrow y''=2{{\sec }^{2}}x\tan x\]
We can see that the second derivation of the trigonometric function \[\tan x\]is over. Now differentiating the second derivative we get third derivative so,
\[\Rightarrow y''=2{{\sec }^{2}}x\tan x\dfrac{dy''}{dx}\]
Now we have to differentiate it though \[uv\] method where the formula is \[{{\left( uv \right)}^{'}}=u'v+uv'\] here,
\[\Rightarrow u={{\sec }^{2}}x\]
\[\Rightarrow v=\tan x\]
Now simplifying we can get the third derivative
\[\Rightarrow y'''=2\left[ 2\sec x\sec x\tan x\tan x+{{\sec }^{2}}x{{\sec }^{2}}x \right]\]
\[\Rightarrow y'''=2\left[ 2{{\sec }^{2}}x{{\tan }^{2}}x+{{\sec }^{4}}x \right]\]
Therefore, the third derivative of the trigonometric function is \[2\left[ 2{{\sec }^{2}}x{{\tan }^{2}}x+{{\sec }^{4}}x \right]\] or \[4{{\sec }^{2}}x{{\tan }^{2}}x+2{{\sec }^{4}}x\].

Note: In this solution we have found the third derivative of the trigonometric function \[\tan x\]. Here we have to remember some differentiation formulas. We have to simplify carefully while differentiating using the \[uv\] method. Students make mistakes while simplifying it.