Question

If ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \left( {\dfrac{\pi }{2}} \right)$, then the value of $\dfrac{{dy}}{{dx}}$ is equal to:
A. $\dfrac{x}{y}$
B. $ - \dfrac{x}{y}$
C. $\dfrac{y}{x}$
D. $ - \dfrac{y}{x}$

Answer 1

Hint: In the given problem, we are required to differentiate both sides of the equation ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \left( {\dfrac{\pi }{2}} \right)$ with respect to x and find the value of $\dfrac{{dy}}{{dx}}$. So, we will have to apply the chain rule of differentiation in the process of differentiation. So, differentiation of ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \left( {\dfrac{\pi }{2}} \right)$ with respect to x will be done layer by layer using the chain rule of differentiation. The derivative of \[{\sin ^{ - 1}}\left( x \right)\] with respect to x must be remembered.

Complete step by step answer:
So, we have, ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \left( {\dfrac{\pi }{2}} \right)$
Differentiating both sides of the equation with respect to x, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\sin }^{ - 1}}x + {{\sin }^{ - 1}}y} \right] = \dfrac{d}{{dx}}\left( {\dfrac{\pi }{2}} \right)$
Now, we know that the derivative of a constant term with respect to x is zero. So, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left[ {{{\sin }^{ - 1}}x} \right] + \dfrac{d}{{dx}}\left[ {{{\sin }^{ - 1}}y} \right] = 0$

Now, we know that the derivative of ${\sin ^{ - 1}}x$ with respect to x is $\dfrac{1}{{\sqrt {1 - {x^2}} }}$. So, we get,
$ \Rightarrow \dfrac{1}{{\sqrt {1 - {x^2}} }} + \dfrac{d}{{dx}}\left[ {{{\sin }^{ - 1}}y} \right] = 0$
We use the chain rule of differentiation $\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$. So, we get,
$ \Rightarrow \dfrac{1}{{\sqrt {1 - {x^2}} }} + \dfrac{1}{{\sqrt {1 - {y^2}} }}\dfrac{{dy}}{{dx}} = 0$
Isolating the differential term $\dfrac{{dy}}{{dx}}$, we get,
$ \Rightarrow \dfrac{1}{{\sqrt {1 - {y^2}} }}\dfrac{{dy}}{{dx}} = - \dfrac{1}{{\sqrt {1 - {x^2}} }}$
Cross multiplying the terms of equation,
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\sqrt {1 - {y^2}} }}{{\sqrt {1 - {x^2}} }} - - - - \left( 1 \right)$

Now, again taking ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = \left( {\dfrac{\pi }{2}} \right)$.
$ \Rightarrow {\sin ^{ - 1}}x = \left( {\dfrac{\pi }{2}} \right) - {\sin ^{ - 1}}y$
We know that the sine inverse and cosine inverse functions are complimentary. So, we get,
$ \Rightarrow {\sin ^{ - 1}}x = {\cos ^{ - 1}}y$
Taking sine on both sides of equation, we get,
$ \Rightarrow \sin \left( {{{\sin }^{ - 1}}x} \right) = \sin \left( {{{\cos }^{ - 1}}y} \right)$
$ \Rightarrow x = \sin \left( {{{\cos }^{ - 1}}y} \right) - - - - \left( 2 \right)$
Let us assume $\left( {{{\cos }^{ - 1}}y} \right)$ as $\theta $.
So, we have, $\theta = \left( {{{\cos }^{ - 1}}y} \right)$.
Taking cosine on both sides of the equation.
$ \Rightarrow \cos \theta = y$

Now, we know that $\sin \theta = \sqrt {1 - {{\cos }^2}\theta } $. So, we get the value of sine as,
$\sin \theta = \sqrt {1 - {y^2}} $
So, $\sin \left( {{{\cos }^{ - 1}}y} \right) = \sqrt {1 - {y^2}} $
Putting this back in equation $\left( 2 \right)$, we get,
$ \Rightarrow x = \sqrt {1 - {y^2}} $
Now, squaring both sides and shifting terms, we get,
$ \Rightarrow {y^2} = 1 - {x^2}$
$ \Rightarrow y = \sqrt {1 - {x^2}} $
Putting these in equation $\left( 1 \right)$, we get,
$ \therefore \dfrac{{dy}}{{dx}} = - \dfrac{x}{y}$
So, the value of $\dfrac{{dy}}{{dx}}$ is $\left( { - \dfrac{x}{y}} \right)$.

Hence, option B is the correct answer.

Note: The derivatives of basic functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer. Care must be taken while doing calculations and simplifying the expressions.