Question

Find the value of the following: $\dfrac{d\left( \tan \left( \dfrac{1}{x} \right) \right)}{dx}$
(a) $-\dfrac{1}{{{x}^{2}}}{{\sec }^{2}}\dfrac{1}{x}$
(b) $-\dfrac{1}{{{x}^{2}}}{{\sin }^{2}}\dfrac{1}{x}$
(c) $-\dfrac{1}{{{x}^{2}}}{{\cos }^{2}}\dfrac{1}{x}$
(d) $-\dfrac{1}{{{x}^{2}}}{{\tan }^{2}}\dfrac{1}{x}$

Answer 1

Hint: Firstly, we need to substitute $\dfrac{1}{x}=t$ so that the given expression will become \[\dfrac{d\left( \tan t \right)}{dx}\]. Then, using the chain rule of the differentiation, we can write the expression as \[\dfrac{d\left( \tan t \right)}{dt}\dfrac{dt}{dx}\]. The differentiation of the function $\tan \theta $ is equal to ${{\sec }^{2}}\theta $, so that it will become ${{\sec }^{2}}t\dfrac{dt}{dx}$. Then on back substituting $t=\dfrac{1}{x}$ the expression will become ${{\sec }^{2}}\dfrac{1}{x}\dfrac{d\left( \dfrac{1}{x} \right)}{dx}$ which can be simplified by using the differentiation of the function ${{x}^{n}}$ which is equal to $n{{x}^{n-1}}$.

Complete step-by-step answer:
Let us consider the expression given in the above question as
\[\Rightarrow E=\dfrac{d\left( \tan \left( \dfrac{1}{x} \right) \right)}{dx}\]
Let us put $\dfrac{1}{x}=t$ so that we can write the above expression as
\[\Rightarrow E=\dfrac{d\left( \tan t \right)}{dx}\]
Now, using the chain rule of the differentiation we can write the above expression as
\[\Rightarrow E=\dfrac{d\left( \tan t \right)}{dt}\dfrac{dt}{dx}\]
Now, we know that the differentiation of the function $\tan \theta $ is equal to ${{\sec }^{2}}\theta $. Therefore, we can put \[\dfrac{d\left( \tan t \right)}{dt}={{\sec }^{2}}t\] in the above expression as
\[\Rightarrow E={{\sec }^{2}}t\dfrac{dt}{dx}\]
Now, substituting $t=\dfrac{1}{x}$ back into the above equation, we can write
\[\Rightarrow E={{\sec }^{2}}\left( \dfrac{1}{x} \right)\dfrac{d\left( \dfrac{1}{x} \right)}{dx}\]
Writing $\dfrac{1}{x}={{x}^{-1}}$ in the above expression we get
\[\Rightarrow E={{\sec }^{2}}\left( \dfrac{1}{x} \right)\dfrac{d\left( {{x}^{-1}} \right)}{dx}\]
Now, we know that the differentiation of the function ${{x}^{n}}$ is equal to \[n{{x}^{n-1}}\]. Therefore, for $n=-1$, the differentiation of the function ${{x}^{-1}}$ will be equal to $-{{x}^{-2}}$ so that the above expression becomes
\[\Rightarrow E={{\sec }^{2}}\left( \dfrac{1}{x} \right)\left( -{{x}^{-2}} \right)\]
Now, using the negative exponent rule we can write the above expression as
\[\begin{align}
  & \Rightarrow E={{\sec }^{2}}\left( \dfrac{1}{x} \right)\left( \dfrac{-1}{{{x}^{2}}} \right) \\
 & \Rightarrow E=\dfrac{-1}{{{x}^{2}}}{{\sec }^{2}}\left( \dfrac{1}{x} \right) \\
\end{align}\]

So, the correct answer is “Option (a)”.

Note: We may forget to apply the chain rule of differentiation and write the expression after differentiating as ${{\sec }^{2}}\dfrac{1}{x}$. To avoid this mistake, we must adopt the method of the substitution, as shown in the above solution.