Question

If \[\sin^{ - 1}x + \sin^{ - 1}y = \dfrac{\pi }{2}\]. Then what is the value of \[\dfrac{{dy}}{{dx}}\]?
A. \[\dfrac{x}{y}\]
B. \[\dfrac{{ - x}}{y}\]
C. \[\dfrac{y}{x}\]
D. \[\dfrac{{ - y}}{x}\]

Answer 1

Hint: Solve the given trigonometric equation u\sing the inverse trigonometric functions. Then differentiate the function with respect to \[x\] and get the required answer.

Formula used:
\[\sin^{ - 1}x + \cos^{ - 1}x = \dfrac{\pi }{2}\]
\[\cos\left( {si{n^{ - 1}}x} \right) = \sqrt {1 - {x^2}} \]
\[\dfrac{d}{{dx}}\left( {\sqrt x } \right) = \dfrac{1}{{2\sqrt x }}\]
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{\left( {n - 1} \right)}}\]

Complete step by step solution:
The given trigonometric equation is \[\sin^{ - 1}x + \sin^{ - 1}y = \dfrac{\pi }{2}\].
Let’s simplify the given equation.
\[\sin^{ - 1}x = \dfrac{\pi }{2} - \sin^{ - 1}y\]
Now apply the formula \[\sin^{ - 1}x + \cos^{ - 1}x = \dfrac{\pi }{2}\] on the right-hand side.
\[\sin^{ - 1}x = \cos^{ - 1}y\]
Apply the formula \[\cos\left( {\sin^{ - 1}x} \right) = \sqrt {1 - {x^2}} \] on the left-hand side.
\[\cos^{ - 1}\sqrt {1 - {x^2}} = \cos^{ - 1}y\]
\[ \Rightarrow \]\[y = \sqrt {1 - {x^2}} \]
Differentiate the above equation with respect to \[x\].
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\sqrt {1 - {x^2}} } \right)\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {1 - {x^2}} }}\dfrac{{dy}}{{dx}}\left( {1 - {x^2}} \right)\] [Since \[\dfrac{d}{{dx}}\left( {\sqrt x } \right) = \dfrac{1}{{2\sqrt x }}\]]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{2\sqrt {1 - {x^2}} }}\left( { - 2x} \right)\] [ Since \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{\left( {n - 1} \right)}}\]]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{{ - x}}{{\sqrt {1 - {x^2}} }}\]
Now substitute \[\sqrt {1 - {x^2}} = y\] in the above equation.
\[\dfrac{{dy}}{{dx}} = \dfrac{{ - x}}{y}\]
Hence the correct option is B.

Note: The inverse trigonometric functions are the inverse functions of basic trigonometric functions sine, cosine, tangent, cosecant, secant, and cotangent. It is used to find the angles with any trigonometric ratio.