Question

The differential coefficient \[{\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)\] with respect to the \[{\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\]
A) 1
B) -1
C) 0
D) None of these

Answer 1

Hint: This question is done using the concept, that is, the substitution of the variables. Variables are substituted by another variable which will help to simplify the given expression.

Formula Used:
1) \[\begin{array}{*{20}{c}}
  {\frac{{dy}}{{dz}}}& = &{\frac{{\left( {\frac{{dy}}{{dx}}} \right)}}{{\left( {\frac{{dz}}{{dx}}} \right)}}}
\end{array}\]
2) \[\begin{array}{*{20}{c}}
  {\tan 2\theta }& = &{\frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}}
\end{array}\]
3) \[\begin{array}{*{20}{c}}
  {\sin 2\theta }& = &{\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}}
\end{array}\]
4) \[\begin{array}{*{20}{c}}
  {\frac{{d({{\tan }^{ - 1}}x)}}{{dx}}}& = &{\frac{1}{{1 + {x^2}}}}
\end{array}\]

Complete step by step Solution:
Whenever differentiation of any expression has to be found with respect to another expression, there is used a formula that is,

\[\begin{array}{*{20}{c}}
  { \Rightarrow \frac{{dy}}{{dz}}}& = &{\frac{{\left( {\frac{{dy}}{{dx}}} \right)}}{{\left( {\frac{{dz}}{{dx}}} \right)}}}
\end{array}\]
Let us assume that y and z are two expressions respectively which is,
\[\begin{array}{*{20}{c}}
  { \Rightarrow y}& = &{{{\tan }^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)}
\end{array}\] and \[\begin{array}{*{20}{c}}
  z& = &{2{{\tan }^{ - 1}}x}
\end{array}\]
Before moving forward, simplify these expressions. For this purpose, substitute x with another variable. Therefore, we can write.
\[\begin{array}{*{20}{c}}
  { \Rightarrow x}& = &{\tan \theta }
\end{array}\]
And
\[\begin{array}{*{20}{c}}
  { \Rightarrow \theta }& = &{{{\tan }^{ - 1}}x}
\end{array}\]
Now, put the value of x in the above expressions. We will get,
\[\begin{array}{*{20}{c}}
  { \Rightarrow y}& = &{{{\tan }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)}
\end{array}\] and \[\begin{array}{*{20}{c}}
  z& = &{{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)}
\end{array}\]
Some trigonometric formulas will now be used to reduce the above expression. So we know that,
\[\begin{array}{*{20}{c}}
  { \Rightarrow \tan 2\theta }& = &{\frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}}
\end{array}\]and \[\begin{array}{*{20}{c}}
  {\sin 2\theta }& = &{\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}}
\end{array}\]
Therefore, we will get.
\[\begin{array}{*{20}{c}}
  { \Rightarrow y}& = &{{{\tan }^{ - 1}}\tan 2\theta }
\end{array}\] and \[\begin{array}{*{20}{c}}
  z& = &{{{\sin }^{ - 1}}\sin 2\theta }
\end{array}\]
And then,
\[\begin{array}{*{20}{c}}
  { \Rightarrow y}& = &{2\theta }
\end{array}\] and \[\begin{array}{*{20}{c}}
  z& = &{2\theta }
\end{array}\]
Now again substitute the value of the \[\theta \]. Therefore, we will get,

\[\begin{array}{*{20}{c}}
  { \Rightarrow y}& = &{2{{\tan }^{ - 1}}x}
\end{array}\] and \[\begin{array}{*{20}{c}}
  z& = &{2{{\tan }^{ - 1}}x}
\end{array}\]
Now differentiate both the expressions with respect to the x. So, we will get it.
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\frac{{dy}}{{dx}}}& = &{2\frac{{d({{\tan }^{ - 1}}x)}}{{dx}}}
\end{array}\]
Now we know that the differentiation of the \[{\tan ^{ - 1}}x\] is,
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\frac{{d({{\tan }^{ - 1}}x)}}{{dx}}}& = &{\frac{1}{{1 + {x^2}}}}
\end{array}\]
Therefore, from the above expression, we will get.
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\frac{{dy}}{{dx}}}& = &{\frac{2}{{1 + {x^2}}}}
\end{array}\]…………….. (a)
And now differentiate the expression z with respect to the x. So, we can write.
\[\begin{array}{*{20}{c}}
  { \Rightarrow \frac{{dz}}{{dx}}}& = &{\frac{2}{{1 + {x^2}}}}
\end{array}\] ………. (b)
According to the question, we will have to find the derivative of one expression with respect to another expression so that we will apply,

\[ \Rightarrow \begin{array}{*{20}{c}}
  {\frac{{dy}}{{dx}}}& = &{\frac{{\left( {\frac{{dy}}{{dx}}} \right)}}{{\left( {\frac{{dz}}{{dx}}} \right)}}}
\end{array}\]
Now from the equation (a) and (b), we will get.
\[\begin{array}{*{20}{c}}
  { \Rightarrow \frac{{dy}}{{dz}}}& = &{\frac{{\frac{2}{{1 + {x^2}}}}}{{\frac{2}{{1 + {x^2}}}}}}
\end{array}\]
Therefore,
\[\begin{array}{*{20}{c}}
  { \Rightarrow \frac{{dy}}{{dz}}}& = &1
\end{array}\]
Now the final answer is 1.

Hence, the correct option is (A).

Note: Use all the basic fundamentals of trigonometry and differentiation to reach the final answer. First, simplify both expressions by substituting the variables and then differentiate both expressions with respect to the x.