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Question: If \[\sin^{- 1}\left( x \right) + \sin^{- 1}\left( y \right) = \dfrac{\pi}{2}\] then what will \[\df...

If sin⁡−1(x)+sin⁡−1(y)=π2\sin^{- 1}\left( x \right) + \sin^{- 1}\left( y \right) = \dfrac{\pi}{2} then what will dydx\dfrac{dy}{dx} be?

Explanation

Solution

In this question , we need to solve the given expression sin⁡−1(x)+sin⁡−1(y)=π2\sin^{- 1}\left( x \right) + \sin^{- 1}\left( y \right) = \dfrac{\pi}{2} and need to find dydx\dfrac{dy}{dx} . First, we have to take the term sin⁡−1(y)\sin^{- 1}(y) to the right hand side and after that we have to use the property of inverse . To find dydx\dfrac{dy}{dx} , we have to convert sin⁡−1\sin^{- 1} in the form of cos⁡−1\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 dydx\dfrac{dy}{dx} .

Formula used :
1. Property of inverse : π2−sin⁡−1(y)=cos⁡−1(y)\dfrac{\pi}{2} - \sin^{- 1}\left( y \right) = \cos^{- 1}(y)
2. sin⁡−1(x)=cos⁡−11–x2\sin^{- 1}(x) = \cos^{- 1}\sqrt{1 – x^{2}}
3. Power rule : d(xn)dx=nxn–1\dfrac{d(x^{n})}{dx} = nx^{n – 1}

Complete step-by-step answer:
Given,
sin⁡−1(x)+sin⁡−1(y)=π2\sin^{- 1}\left( x \right) + \sin^{- 1}\left( y \right) = \dfrac{\pi}{2}
Here we need to find dydx\dfrac{dy}{dx}
First, we have to take the term sin⁡−1(y)\sin^{- 1}\left( y \right) to the right hand side.
Given,
sin⁡−1(x)+sin⁡−1(y)=π2\sin^{- 1}\left( x \right) + \sin^{- 1}\left( y \right) = \dfrac{\pi}{2}
On taking the term sin⁡−1(y)\sin^{- 1}\left( y \right) to the right hand side,
We get,
⇒ sin⁡−1(x)=π2−sin⁡−1(y)\Rightarrow \ \sin^{- 1}\left( x \right) = \dfrac{\pi}{2} - \sin^{- 1}\left( y \right)
Now using the property of inverse,
We get,
⇒ sin⁡−1(x)=cos⁡−1(y)\Rightarrow \ \sin^{- 1}\left( x \right) = \cos^{- 1}(y)
In order to convert sin⁡−1\sin^{- 1} in the form of cos⁡−1\cos^{- 1} which can be converted with the help of the formula sin⁡−1(x)=cos⁡−11–x2\sin^{- 1}(x) = \cos^{- 1}\sqrt{1 – x^{2}} ,
⇒ cos⁡−11–x2=cos⁡−1(y)\Rightarrow \ \cos^{- 1}\sqrt{1 – x^{2}} = \cos^{- 1}\left( y \right)
Now on taking cos on both sides,
We get,
⇒ cos⁡(cos⁡−11–x2)=cos⁡(cos⁡−1(y))\Rightarrow \ \cos \left( \cos^{- 1}\sqrt{1 – x^{2}} \right) = \cos\left( \cos^{- 1}\left( y \right) \right)
On simplifying,
We get,
⇒1–x2=y\Rightarrow \sqrt{1 – x^{2}} = y ••• (1)
On rewriting,
We get,
y=(1–x2)12y = \left( 1 – x^{2} \right)^{\dfrac{1}{2}}
Now on differentiating both sides,
We get,
dydx=12(1–x2)(12–1)×(−2x)\dfrac{dy}{dx} = \dfrac{1}{2}\left( 1 – x^{2} \right)^{(\dfrac{1}{2} – 1)} \times \left( - 2x \right)
On simplifying,
We get,
dydx=−2x2(1–x2)−12\dfrac{dy}{dx} = - \dfrac{2x}{2}\left( 1 – x^{2} \right)^{- \dfrac{1}{2}}
On further simplifying,
We get,
dydx=−x(1–x2)−12\dfrac{dy}{dx} = - \dfrac{x}{\left( 1 – x^{2} \right)^{- \dfrac{1}{2}}}
We can rewrite (a)12\left( a \right)^{\dfrac{1}{2}} as a\sqrt{a} ,
Thus we get,
dydx=−x1–x2\dfrac{dy}{dx} = - \dfrac{x}{\sqrt{1 – x^{2}}}
From equation (1) y=1–x2y = \sqrt{1 – x^{2}} ,
We get,
dydx=−xy\dfrac{dy}{dx} = - \dfrac{x}{y}
Thus we get the value of dydx\dfrac{dy}{dx} is −xy- \dfrac{x}{y}
Final answer :
The value of dydx\dfrac{dy}{dx} is −xy- \dfrac{x}{y}

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