If \[\sin^{- 1}\left( x \right) + \sin^{- 1}\left( y \right) = \dfrac{\pi}{2}\] then what will \[\dfrac{dy}{dx}\] be?

Hint: In this question , we need to solve the given expression \[\sin^{- 1}\left( x \right) + \sin^{- 1}\left( y \right) = \dfrac{\pi}{2}\] and need to find \[\dfrac{dy}{dx}\] . First, we have to take the term \[\sin^{- 1}(y)\] to the right hand side and after that we have to use the property of inverse . To find \[\dfrac{dy}{dx}\] , we have to convert \[\sin^{- 1}\] in the form of \[\cos^{- 1}\] which can be converted with the help of the trigonometry formula . Now we can differentiate the expression by using the power rule. After doing the differentiation, then we need to do some rearrangements of terms and hence we find \[\dfrac{dy}{dx}\] .

Formula used :
1. Property of inverse : \[\dfrac{\pi}{2} - \sin^{- 1}\left( y \right) = \cos^{- 1}(y)\]
2. \[\sin^{- 1}(x) = \cos^{- 1}\sqrt{1 – x^{2}}\]
3. Power rule : \[\dfrac{d(x^{n})}{dx} = nx^{n – 1}\]

Complete step-by-step answer:
Given,
\[\sin^{- 1}\left( x \right) + \sin^{- 1}\left( y \right) = \dfrac{\pi}{2}\]
Here we need to find \[\dfrac{dy}{dx}\]
First, we have to take the term \[\sin^{- 1}\left( y \right)\] to the right hand side.
Given,
\[\sin^{- 1}\left( x \right) + \sin^{- 1}\left( y \right) = \dfrac{\pi}{2}\]
On taking the term \[\sin^{- 1}\left( y \right)\] to the right hand side,
We get,
\[\Rightarrow \ \sin^{- 1}\left( x \right) = \dfrac{\pi}{2} - \sin^{- 1}\left( y \right)\]
Now using the property of inverse,
We get,
\[\Rightarrow \ \sin^{- 1}\left( x \right) = \cos^{- 1}(y)\]
In order to convert \[\sin^{- 1}\] in the form of \[\cos^{- 1}\] which can be converted with the help of the formula \[\sin^{- 1}(x) = \cos^{- 1}\sqrt{1 – x^{2}}\] ,
\[\Rightarrow \ \cos^{- 1}\sqrt{1 – x^{2}} = \cos^{- 1}\left( y \right)\]
Now on taking cos on both sides,
We get,
\[\Rightarrow \ \cos \left( \cos^{- 1}\sqrt{1 – x^{2}} \right) = \cos\left( \cos^{- 1}\left( y \right) \right)\]
On simplifying,
We get,
\[\Rightarrow \sqrt{1 – x^{2}} = y\] ••• (1)
On rewriting,
We get,
\[y = \left( 1 – x^{2} \right)^{\dfrac{1}{2}}\]
Now on differentiating both sides,
We get,
\[\dfrac{dy}{dx} = \dfrac{1}{2}\left( 1 – x^{2} \right)^{(\dfrac{1}{2} – 1)} \times \left( - 2x \right)\]
On simplifying,
We get,
\[\dfrac{dy}{dx} = - \dfrac{2x}{2}\left( 1 – x^{2} \right)^{- \dfrac{1}{2}}\]
On further simplifying,
We get,
\[\dfrac{dy}{dx} = - \dfrac{x}{\left( 1 – x^{2} \right)^{- \dfrac{1}{2}}}\]
We can rewrite \[\left( a \right)^{\dfrac{1}{2}}\] as \[\sqrt{a}\] ,
Thus we get,
\[\dfrac{dy}{dx} = - \dfrac{x}{\sqrt{1 – x^{2}}}\]
From equation (1) \[y = \sqrt{1 – x^{2}} \] ,
We get,
\[\dfrac{dy}{dx} = - \dfrac{x}{y}\]
Thus we get the value of \[\dfrac{dy}{dx}\] is \[- \dfrac{x}{y}\]
Final answer :
The value of \[\dfrac{dy}{dx}\] is \[- \dfrac{x}{y}\]

Note: To find \[\dfrac{dy}{dx}\] , it is not necessary that we have to take \[\sin^{- 1}\left( y \right)\] to the right hand side of the expression we can also take \[\sin^{- 1}\left( x \right)\] to the right hand side of the expression. If we take \[\sin^{- 1}\left( y \right)\] to the other side of the expression we will get the answer in form of \[x\] or if we take \[\sin^{- 1}\left( x \right)\] to the right hand side and on solving we will obtain the answer in form of \[y\] . Also, while differentiating we should be careful in using the power rule \[\dfrac{dx^{n}}{dx} = nx^{n – 1}\] , a simple error that may happen while calculating.