Question

Differential coefficient of \[{\sin ^{ - 1}}(\dfrac{{1 - x}}{{1 + x}})\] w.r.t \[\sqrt x \] is:
A. \[\dfrac{1}{{2\sqrt x }}\]
B. \[\dfrac{{\sqrt x }}{{\sqrt {1 - x} }}\]
C. \[1\]
D. \[ - \dfrac{2}{{1 + x}}\]

Answer 1

Hint:This question can be solved by changing the variable of the function.The method of substitution will simplify the question, and therefore, we can then easily differentiate the entire function.

Formula used:
The formulas involved in this question are:
\[\dfrac{{d\tan (\theta )}}{{d\theta }} = {\sec ^2}(\theta )\],
\[\Rightarrow {\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}\], for all values of \[\theta \]
\[\Rightarrow {\sec ^2}(\theta ) - {\tan ^2}(\theta ) = 1\], for all values of \[\theta \]
\[\Rightarrow {\cos ^2}(\theta ) - {\sin ^2}(\theta ) = \cos (2\theta )\], for all values of \[\theta \]

Complete step by step answer:
 Let us start the question with what we are asked to find, i.e.,
\[ \Rightarrow \dfrac{{d{{\sin }^{ - 1}}(\dfrac{{1 - x}}{{1 + x}})}}{{d\sqrt x }}\]
Now, to simplify the derivative, let's make a substitution as shown below
\[ \Rightarrow x = {\tan ^2}(\theta )\]
The benefit of using this substitution is that it simplifies the function \[{\sin ^{ - 1}}\] as well \[\sqrt x \] and makes them linear.
\[ \Rightarrow {\sin ^{ - 1}}(\dfrac{{1 - x}}{{1 + x}}) = {\sin ^{ - 1}}(\dfrac{{1 - {{\tan }^2}(\theta )}}{{1 + {{\tan }^2}(\theta )}})\]
Now, by solving the angle for \[{\sin ^{ - 1}}\] we get,
\[ \Rightarrow {\sin ^{ - 1}}(\dfrac{{1 - x}}{{1 + x}}) = {\sin ^{ - 1}}(\dfrac{{{{\cos }^2}(\theta ) - {{\sin }^2}(\theta )}}{{{{\cos }^2}(\theta ) + {{\sin }^2}(\theta )}})\]

Now, let us apply the trigonometric identities \[{\cos ^2}(\theta ) - {\sin ^2}(\theta ) = \cos (2\theta )\] and \[{\cos ^2}(\theta ) + {\sin ^2}(\theta ) = 1\]to get,
\[ \Rightarrow {\sin ^{ - 1}}(\dfrac{{1 - x}}{{1 + x}}) = {\sin ^{ - 1}}(\dfrac{{\cos (2\theta )}}{1})\]
Now, by using, \[{\sin ^{ - 1}}\theta + {\cos ^{ - 1}}\theta = \dfrac{\pi }{2}\] we get,
\[ \Rightarrow {\sin ^{ - 1}}(\dfrac{{1 - x}}{{1 + x}}) = \dfrac{\pi }{2} - (2\theta )\]
Thus, we have successfully simplified \[{\sin ^{ - 1}}\]in terms of \[\theta \].
Now, let us replace this value in the differential equation as shown below,
\[ \Rightarrow \dfrac{{d{{\sin }^{ - 1}}(\dfrac{{1 - x}}{{1 + x}})}}{{d\sqrt x }} = \dfrac{{d(\dfrac{\pi }{2} - (2\theta ))}}{{d\tan (\theta )}}\]
Now, by multiplying and dividing by \[d\theta \]on the right-hand side of the equation we get,
\[ \Rightarrow \dfrac{{d(\dfrac{\pi }{2} - (2\theta ))}}{{d\tan (\theta )}} = \dfrac{{d(\dfrac{\pi }{2} - (2\theta ))}}{{d\theta }} \times \dfrac{{d\theta }}{{d\tan (\theta )}}\]

Now, as the differentiation are pretty elementary, we directly substitute their values to get,
\[ \Rightarrow \dfrac{{d(\dfrac{\pi }{2} - (2\theta ))}}{{d\theta }} \times \dfrac{{d\theta }}{{d\tan (\theta )}} = - 2 \times \dfrac{1}{{{{\sec }^2}(\theta )}}\]
Now, let us apply the trigonometric identity \[{\sec ^2}(\theta ) - {\tan ^2}(\theta ) = 1\] to get,
\[ \Rightarrow - 2 \times \dfrac{1}{{1 + {{\tan }^2}(\theta )}}\]
Finally, replacing the value of \[{\tan ^2}(\theta )\] as \[x\]we get,
\[ \Rightarrow - 2 \times \dfrac{1}{{1 + {{\tan }^2}(\theta )}} = - \dfrac{2}{{1 + x}}\]
\[ \therefore \dfrac{{d{{\sin }^{ - 1}}(\dfrac{{1 - x}}{{1 + x}})}}{{d\sqrt x }} = - \dfrac{2}{{1 + x}}\]

Therefore, option D is the correct answer.

Note: This question involves multiple concepts like trigonometry, differentiation. One should be well versed with these topics to solve this question. Here a unique substitution is made; try to practice similar questions so that these substitutions become elementary. Calculation mistakes are possible in this question, so try to avoid them and be sure of the final answer.