Question

Find the differential coefficient of ${{\tan }^{-1}}x$?

Answer 1

Hint: First of all equate ${{\tan }^{-1}}x$ with $f\left( x \right)$ which will look as ${{\tan }^{-1}}x=f\left( x \right)$. Substitute x as $\tan \theta $ in this equation which will look like $f\left( \tan \theta \right)=\theta $. Now, take the derivative with respect to $\theta $ on both the sides which will give $f'\left( \tan \theta \right){{\sec }^{2}}\theta =1$ after that write ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ then substitute $\tan \theta $ as x and hence, you will get the differential coefficient of ${{\tan }^{-1}}x$.

Complete step by step solution:
As ${{\tan }^{-1}}x$ is the function of x so we can equate ${{\tan }^{-1}}x$ to $f\left( x \right)$.
${{\tan }^{-1}}x=f\left( x \right)$
Now, substituting $x=\tan \theta $ in the above equation we get,
${{\tan }^{-1}}\left( \tan \theta \right)=f\left( \tan \theta \right)$………….. Eq. (1)
We know that from the algebra that when a term and its inverse are written side by side then the term and its inverse vanishes to 1 so ${{\tan }^{-1}}\left( \tan \theta \right)$ can be written as:
${{\tan }^{-1}}\left( \tan \theta \right)=\theta $
Substituting the above relation in eq. (1) we get,
$\theta =f\left( \tan \theta \right)$
Differentiating on both the sides with respect to $\theta $ we get,
$1=f'\left( \tan \theta \right){{\sec }^{2}}\theta $………… Eq. (2)
In the above differentiation, derivative of $\theta $ with respect to $\theta $ is 1 and we have differentiated $f\left( \tan \theta \right)$ with respect to $\theta $ by chain rule in which first we have differentiated $f\left( \tan \theta \right)$ which is $f'\left( \tan \theta \right)$ and multiplied with the derivative of $\tan \theta $ which is ${{\sec }^{2}}\theta $.
Dividing ${{\sec }^{2}}\theta $ on both the sides of the eq. (2) we get,
$f'\left( \tan \theta \right)=\dfrac{1}{{{\sec }^{2}}\theta }$……….. Eq. (3)
We know from the trigonometric identities that:
${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $
Substituting the above relation in eq. (3) we get,
$f'\left( \tan \theta \right)=\dfrac{1}{1+{{\tan }^{2}}\theta }$
In the start of the solution, we have assumed that $x=\tan \theta $ so substituting x in place of $\tan \theta $ we get,
$f'\left( x \right)=\dfrac{1}{1+{{x}^{2}}}$………. Eq. (4)
As we have shown above that ${{\tan }^{-1}}x=f\left( x \right)$ so taking derivative on both the sides will give:
$f'\left( x \right)=\dfrac{d{{\tan }^{-1}}x}{dx}$ ……….. Eq. (5)
Comparing eq. (4) and eq. (5) we get,
$\dfrac{d{{\tan }^{-1}}x}{dx}=\dfrac{1}{1+{{x}^{2}}}$
Hence, the differential coefficient of ${{\tan }^{-1}}x$ is equal to $\dfrac{1}{1+{{x}^{2}}}$.

Note: The question demands a good understanding of trigonometric identities and properties of the inverse and the derivative methods like here, we have used chain rule.
The trigonometric knowledge that we require in this problem is ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $.
The point to remember in this problem is to remember the differential coefficient of ${{\tan }^{-1}}x$. If you know the derivative of ${{\tan }^{-1}}x$ then you will save time in the competitive exams because in competitive exams, you will quite often find the application of derivative of ${{\tan }^{-1}}x$.