Question

Solve the following:
$\dfrac{d}{{dx}}[{\tan ^{ - 1}}\dfrac{{(a - x)}}{{(1 + ax)}}] =$ ?
$A)\dfrac{1}{{1 + x{}^2}}$
$B)\dfrac{{ - 1}}{{1 + x{}^2}}$
$C)\dfrac{a}{{1 + x{}^2}}$
$D)$ None of these

Answer 1

First, we need to analyze the given information which is in the trigonometric form.
Here we are asked to calculate the differentiation of ${\tan ^{ - 1}}\dfrac{{(a - x)}}{{(1 + ax)}}$ , so we need to know the formulas in $\tan$ (tangent) also.
In differentiation, the derivative of $x$ raised to the power is denoted by $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ .
Formula used:
$\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ and $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
$\dfrac{d}{{dx}}({\tan ^{ - 1}}x) = \dfrac{1}{{1 + {x^2}}}$
${\tan ^{ - 1}}(\tan ) = 1$ (Which is the inverse function of original function gets one)

Complete step by step answer:
Since from the given that, we have $\dfrac{d}{{dx}}[{\tan ^{ - 1}}\dfrac{{(a - x)}}{{(1 + ax)}}]$
Let us take the function tangent into the form of $y = [{\tan ^{ - 1}}\dfrac{{(a - x)}}{{(1 + ax)}}]$
Now assume $a = \tan \alpha ,x = \tan \theta$ then substitute the values into the above equation we get, $y = [{\tan ^{ - 1}}\dfrac{{(a - x)}}{{(1 + ax)}}] \Rightarrow [{\tan ^{ - 1}}\dfrac{{(\tan \alpha - \tan \theta )}}{{(1 + \tan \alpha .\tan \theta )}}]$
Now from the tangent formula, we know that, $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$ and substituting this in the above equation we get,
$y = [{\tan ^{ - 1}}\dfrac{{(\tan \alpha - \tan \theta )}}{{(1 + \tan \alpha .\tan \theta )}}] \Rightarrow {\tan ^{ - 1}}[\tan (\alpha - \theta )]$ where $\tan (\alpha - \theta ) = \dfrac{{\tan \alpha - \tan \theta }}{{1 + \tan \alpha \tan \theta }}$
Again, from the given formula we know that, ${\tan ^{ - 1}}(\tan ) = 1$
Substituting these value into the above equation we get, $y = {\tan ^{ - 1}}[\tan (\alpha - \theta )] \Rightarrow (\alpha - \theta )$
So, we converted the original given form of a tangent as $y = (\alpha - \theta )$
But since in the beginning, we assumed that $a = \tan \alpha ,x = \tan \theta$ and this can be rewritten as $a = \tan \alpha ,x = \tan \theta \Rightarrow \alpha = {\tan ^{ - 1}}a,\theta = {\tan ^{ - 1}}x$ (where these functions are the inverse image to the given functions)
Now substitute the converted values $\alpha = {\tan ^{ - 1}}a,\theta = {\tan ^{ - 1}}x$ in the above values $y = (\alpha - \theta )$ then we get, $y = (\alpha - \theta ) \Rightarrow {\tan ^{ - 1}}a - {\tan ^{ - 1}}x$
Finally, we will take the differentiation with respect to x (which is the requirement)
Thus, we get, $\dfrac{{d(y)}}{{dx}} = \dfrac{d}{{dx}}({\tan ^{ - 1}}a - {\tan ^{ - 1}}x)$
Now giving the derivative inside the tangent function we get, $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\tan ^{ - 1}}a - \dfrac{d}{{dx}}{\tan ^{ - 1}}x$
Since we know that from the formula of a tangent, $\dfrac{d}{{dx}}({\tan ^{ - 1}}x) = \dfrac{1}{{1 + {x^2}}}$ (with respect to x) and $\dfrac{d}{{dx}}{\tan ^{ - 1}}a = 0$ (because this is a constant function, the function is in with constant value $a$ so if we differentiate the constant, we only get zero)
Hence substitute both values of $\dfrac{d}{{dx}}({\tan ^{ - 1}}x) = \dfrac{1}{{1 + {x^2}}}$ and $\dfrac{d}{{dx}}{\tan ^{ - 1}}a = 0$ in $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\tan ^{ - 1}}a - \dfrac{d}{{dx}}{\tan ^{ - 1}}x$ .
Thus, we get, $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\tan ^{ - 1}}a - \dfrac{d}{{dx}}{\tan ^{ - 1}}x \Rightarrow 0 - \dfrac{1}{{1 + {x^2}}}$
Therefore, finally, we get $\dfrac{{dy}}{{dx}} = 0 - \dfrac{1}{{1 + {x^2}}} \Rightarrow - \dfrac{1}{{1 + {x^2}}}$

So, the correct answer is “Option B”.

Note:
Since if we substitute the wrong formula for the tangent, that is $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ then there is the possibility to get the option as $A)\dfrac{1}{{1 + x{}^2}}$ but which is the completely wrong answer. So be careful at the tangent formula which is in the solution we only required $y = {\tan ^{ - 1}}\dfrac{{(\tan \alpha - \tan \theta )}}{{(1 + \tan \alpha .\tan \theta )}}$ and in the numerator, it is a negative sign so we need to apply the formula as $\tan (A - B) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$