Question

Match the following columns:

Answer 1

Hint: To match the columns, simplify the functions given on the right hand side using the properties of inverse functions in the given range and domain and then differentiate them to get the exact values and then match them with the right hand side.

Complete step-by-step answer:

We have three different kinds of functions on the right side of the table. We want to find their derivative. We will begin by simplifying the given functions in their possible domain and range and then finding their first derivative.
We have the function $y={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$. We know that the domain of $y={{\sin }^{-1}}x$ is $\left[ -1,1 \right]$ and its range is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.
Thus, by substituting $x=\tan \theta$ we get
$y={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x$ if $\left| x \right|\le 1$
$=\pi -2{{\tan }^{-1}}x$ if $x>1$
$=-\left( \pi +2{{\tan }^{-1}}x \right)$ if $x<-1$
We know that the first derivative of $y=a{{\tan }^{-1}}x+b$ is $\dfrac{dy}{dx}=\dfrac{a}{1+{{x}^{2}}}$ .
If $\left| x \right|<1$, we have $y={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x$, thus $\dfrac{dy}{dx}=\dfrac{2}{1+{{x}^{2}}}$.
If $\left| x \right|>1$, we have $y={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\pm \pi -2{{\tan }^{-1}}x$, thus $\dfrac{dy}{dx}=\dfrac{-2}{1+{{x}^{2}}}$.
Thus, for (a) the correct options are (p), (s).
We have the function $y={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$. We know that the domain of $y={{\cos }^{-1}}x$ is $\left[ -1,1 \right]$ and its range is $\left[ 0,\pi \right]$.
Thus, by substituting $x=\tan \theta$ we get
$y={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x$ if $x\ge 0$
$=-2{{\tan }^{-1}}x$ if $x<0$
We know that the first derivative of $y=a{{\tan }^{-1}}x+b$ is $\dfrac{dy}{dx}=\dfrac{a}{1+{{x}^{2}}}$ .
If $x<0$, we have $y={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=-2{{\tan }^{-1}}x$, thus $\dfrac{dy}{dx}=\dfrac{-2}{1+{{x}^{2}}}$.
If $x\ge 0$, we have $y={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x$, thus $\dfrac{dy}{dx}=\dfrac{2}{1+{{x}^{2}}}$.
Thus, for (b) the correct option is (r).
We have the function $y={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$. We know that the domain of $y={{\tan }^{-1}}x$ is $\mathbb{R}$ and its range is $\left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$.
Thus, by substituting $x=\tan \theta$ we get
$y={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)=2{{\tan }^{-1}}x$ if $\left| x \right|<1$
$=-\pi +2{{\tan }^{-1}}x$ if $x>1$
$=\pi +2{{\tan }^{-1}}x$ if $x<-1$
We know that the first derivative of $y=a{{\tan }^{-1}}x+b$ is $\dfrac{dy}{dx}=\dfrac{a}{1+{{x}^{2}}}$ .
If $\left| x \right|<1$, we have $y={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)=2{{\tan }^{-1}}x$, thus $\dfrac{dy}{dx}=\dfrac{2}{1+{{x}^{2}}}$.
If $\left| x \right|>1$, we have $y={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)=\pm \pi +2{{\tan }^{-1}}x$, thus $\dfrac{dy}{dx}=\dfrac{2}{1+{{x}^{2}}}$.
Also, the function $y={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ is non-existent for $\left| x \right|=1$.
Thus, for (c) the correct options are (p), (q) and (t).

Note: It’s necessary to keep in mind the possible domain and range of the given inverse functions. The functions show different behaviour in different values of domain and range. If we don’t keep the domain and range in mind, we will get an incorrect answer.