Question

If we have $y=\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$, then prove that $\left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}-xy=1$.

Answer 1

Hint: In this question, we need to use the quotient rule of differentiation to prove the required result. First, consider ‘u’ and ‘v’ and then solve accordingly to the quotient rule, which is, $\dfrac{dy}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ and then use various other differentiation formulas including the chain rule and some basic operations of mathematics to prove the required result.

Complete step-by-step solution:
We have been given, $y=\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$
The above expression is in the form of $\dfrac{u}{v}$, therefore, we can use the quotient rule in differentiation for $y=\dfrac{u}{v}$. Let us compare, $y=\dfrac{u}{v}$ to $y=\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$, so we get, $u={{\sin }^{-1}}x,v=\sqrt{1-{{x}^{2}}}$.
Now, according to the quotient rule, we know that,
$\dfrac{dy}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$
On differentiating the equation with respect to x, we get,
$\dfrac{dy}{dx}=\dfrac{\left( \sqrt{1-{{x}^{2}}} \right)\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)-\left( {{\sin }^{-1}}x \right)\dfrac{d}{dx}\left( \sqrt{1-{{x}^{2}}} \right)}{{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}$
We know that,
$\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ and,
$\begin{align}
  & \dfrac{d}{dx}\left( \sqrt{1-{{x}^{2}}} \right)=\dfrac{1}{2\sqrt{1-{{x}^{2}}}}.\dfrac{d}{dx}\left( 1-{{x}^{2}} \right)\text{ }\left[ \text{By applying the chain rule} \right] \\
 & =\dfrac{1}{2\sqrt{1-{{x}^{2}}}}\left( -2x \right) \\
\end{align}$
Now, on substituting the above values in the $\dfrac{dy}{dx}$, we get,
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{\left( \sqrt{1-{{x}^{2}}} \right)\dfrac{1}{\left( \sqrt{1-{{x}^{2}}} \right)}-\left( {{\sin }^{-1}}x \right)\left( \dfrac{1}{2\sqrt{1-{{x}^{2}}}} \right)\left( -2x \right)}{1-{{x}^{2}}} \\
 & \dfrac{dy}{dx}=\dfrac{1+\left( {{\sin }^{-1}}x \right)\left( \dfrac{1}{2\sqrt{1-{{x}^{2}}}} \right)\left( 2x \right)}{1-{{x}^{2}}} \\
\end{align}\]
After cancelling the number 2, in the numerator, and on cross multiplying the above equation we get,
\[\left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}=1+\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}x\]
From the question, we know that, $y=\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}$.
Therefore, \[\left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}=1+xy\]
Subtracting xy on both the sides, we get,
\[\begin{align}
  & \left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}-xy=1+xy-xy \\
 & \left( 1-{{x}^{2}} \right)\dfrac{dy}{dx}-xy=1 \\
\end{align}\]
Hence proved.

Note: The quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions. Apart from the quotient rule, there is also a product rule when two differentiable functions are multiplied to each other. The product rule is as follows, $\dfrac{d}{dx}\left( UV \right)=\dfrac{du}{dx}v+u\dfrac{dv}{dx}$. Differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another variable. Another name for the differentiation is derivative. It is used as a derivative of a particular function.