Question

$\dfrac{{d[{{\tan }^{ - 1}}\dfrac{{(a - x)}}{{1 + ax}}]}}{{dx}} = $
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: This question is from the chapter, named Trigonometry. Apply the formula of inverse tan to reduce the trigonometric expression. Use all the basic formulas of trigonometry to solve the trigonometric expression. After that apply the differentiation formula that will help to reduce the expression.

Formula Used:
\[{\tan ^{ - 1}}a - {\tan ^{ - 1}}b = {\tan ^{ - 1}}\dfrac{{(a - b)}}{{1 + ab}}\]

Complete step by step Solution:
There are certain steps involved to solve these kinds of questions. A certain producer should be followed to simplify these kinds of trigonometric expressions.
Our main purpose is to simplify the above expression as much as we can. So, to do that, we will have to use all the basic fundamentals of trigonometry.
 In addition to two different tan inverse functions, the formula will be represented as below.
\[{\tan ^{ - 1}}a - {\tan ^{ - 1}}b = {\tan ^{ - 1}}\dfrac{{(a - b)}}{{1 + ab}}\]

This is also known as the additional formula for the inverse of tan. It is derived from the addition of two inverses of tan.
Using our question we get,
\[y = {\tan ^{ - 1}}\dfrac{{(a - x)}}{{1 + ax}} = {\tan ^{ - 1}}a - {\tan ^{ - 1}}x\]
Now differentiating y with respect to x we get,

$\dfrac{{dy}}{{dx}} = 0 - \dfrac{1}{{1 + {x^2}}}$ [since $\dfrac{{d{{\tan }^{ - 1}}x}}{{dx}} = \dfrac{1}{{1 + {x^2}}}$ and also differentiation of a constant is zero]

$\dfrac{{d[y]}}{{dx}} = \dfrac{{ - 1}}{{1 + {x^2}}}$

Hence, the correct option is B.

Note: Use all the basic trigonometric formulas to reduce the expression. Also, use the differentiation of trigonometric functions and apply these formulas until the expressions get simple. After that use trigonometric ratios. All the formulas that you are going to apply, should be in such a manner that there are no errors in the solution.