Question

Find the derivative $\dfrac{d}{dx}\left[ {{\tan }^{-1}}\left( \dfrac{ax-b}{bx+a} \right) \right]$?
A. $\dfrac{1}{1+{{x}^{2}}}-\dfrac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}$
B. $\dfrac{-1}{1+{{x}^{2}}}-\dfrac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}$
C. $\dfrac{1}{1+{{x}^{2}}}+\dfrac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}$
D. None of these

Answer 1

We first define the chain rule and how the differentiation of composite function works. We differentiate the main function with respect to the intermediate function and then take differentiation of the intermediate function with respect to $x$. We take multiplication of these two different differentiated values.

Complete step by step answer:
We differentiate the given function $f\left( x \right)={{\tan }^{-1}}\left( \dfrac{ax-b}{bx+a} \right)$ with respect to $x$ using the chain rule.
Here we have a composite function where the main function is $g\left( x \right)={{\tan }^{-1}}x$ and the other function is $h\left( x \right)=\dfrac{ax-b}{bx+a}$.
We have $goh\left( x \right)=g\left( \dfrac{ax-b}{bx+a} \right)={{\tan }^{-1}}\left( \dfrac{ax-b}{bx+a} \right)$. We take this as our $f\left( x \right)={{\tan }^{-1}}\left( \dfrac{ax-b}{bx+a} \right)$.
We need to find the value of $\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ {{\tan }^{-1}}\left( \dfrac{ax-b}{bx+a} \right) \right]$. We know $f\left( x \right)=goh\left( x \right)$.
Differentiating $f\left( x \right)=goh\left( x \right)$, we get
$\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ goh\left( x \right) \right]=\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}={{g}^{'}}\left[ h\left( x \right) \right]{{h}^{'}}\left( x \right)$.
The above-mentioned rule is the chain rule.
The chain rule allows to differentiate with respect to the function $h\left( x \right)$ instead of $x$ and after that we need to take the differentiated form of $h\left( x \right)$ with respect to $x$.
For the function $f\left( x \right)={{\tan }^{-1}}\left( \dfrac{ax-b}{bx+a} \right)$, we take differentiation of $f\left( x \right)={{\tan }^{-1}}\left( \dfrac{ax-b}{bx+a} \right)$ with respect to the function $h\left( x \right)=\dfrac{ax-b}{bx+a}$ instead of $x$ and after that we need to take the differentiated form of $h\left( x \right)=\dfrac{ax-b}{bx+a}$ with respect to $x$.
We know that differentiation of $g\left( x \right)={{\tan }^{-1}}x$ is ${{g}^{'}}\left( x \right)=\dfrac{1}{1+{{x}^{2}}}$ and differentiation of $h\left( x \right)=\dfrac{ax-b}{bx+a}$ is ${{h}^{'}}\left( x \right)=\dfrac{a\left( bx+a \right)-b\left( ax-b \right)}{{{\left( bx+a \right)}^{2}}}=\dfrac{{{a}^{2}}+{{b}^{2}}}{{{\left( bx+a \right)}^{2}}}$.
$\Rightarrow \dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{d\left[ \dfrac{ax-b}{bx+a} \right]}\left[ {{\tan }^{-1}}\left( \dfrac{ax-b}{bx+a} \right) \right]\times \dfrac{d\left[ \dfrac{ax-b}{bx+a} \right]}{dx}$
We place the values of the differentiations and get

& \Rightarrow \dfrac{d}{dx}\left[ f\left( x \right) \right] \\\ & =\dfrac{1}{1+{{\left( \dfrac{ax-b}{bx+a} \right)}^{2}}}\left[ \dfrac{{{a}^{2}}+{{b}^{2}}}{{{\left( bx+a \right)}^{2}}} \right] \\\ & =\dfrac{{{a}^{2}}+{{b}^{2}}}{{{\left( bx+a \right)}^{2}}+{{\left( ax-b \right)}^{2}}} \\\ & =\dfrac{{{a}^{2}}+{{b}^{2}}}{{{x}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)+\left( {{a}^{2}}+{{b}^{2}} \right)} \\\ & =\dfrac{{{a}^{2}}+{{b}^{2}}}{\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{x}^{2}}+1 \right)} \\\ & =\dfrac{1}{\left( {{x}^{2}}+1 \right)} \\\ \end{aligned}$$ **Therefore, the correct option is option (D).** **Note:** We need remember that in the chain rule $$\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}$$, we aren’t cancelling out the part $$d\left[ h\left( x \right) \right]$$. Cancelation of the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to differentiate.