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

Hint: 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}} $