Q.If y = e^{m\cos^{-1}x}, then prove that (1 - x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} - m^2 y = 0. OR If x = a(\cos\theta + \theta\sin\theta) and y = a(\sin\theta - \theta\cos\theta), then find \dfrac{dy}{dx}.

The problem asks us to find the value of the expression $(1 - x^2)\frac{d^2y}{dx^2}$ given that $y = \sin^{-1} x$. This requires us to calculate both the first derivative ($\frac{dy}{dx}$) and the second derivative ($\frac{d^2y}{dx^2}$) of $y$ with respect to $x$. Once we have these, we will substitute them into the given expression and simplify.

A key strategy in problems involving higher-order derivatives of inverse trigonometric functions is to simplify the first derivative expression before differentiating it again. This often involves eliminating square roots or fractions, which makes the subsequent differentiation much cleaner and less prone to errors.

1. Find the first derivative, $\frac{dy}{dx}$.

   We are given the function $y = \sin^{-1} x$.

   The standard derivative of $\sin^{-1} x$ with respect to $x$ is:

   If $y = \sin^{-1} x$, then $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$.

   So, our first derivative is:

   $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$

2. Prepare for the second derivative by simplifying the first derivative expression.

   To make the calculation of the second derivative easier, we can rearrange the expression for $\frac{dy}{dx}$ to remove the square root from the denominator. This is a common and effective technique.

   Multiply both sides by $\sqrt{1 - x^2}$:

   $\sqrt{1 - x^2} \frac{dy}{dx} = 1$

   Now, to eliminate the square root entirely, we square both sides of the equation:

   $\left(\sqrt{1 - x^2}\right)^2 \left(\frac{dy}{dx}\right)^2 = 1^2$

   $(1 - x^2) \left(\frac{dy}{dx}\right)^2 = 1$

   This form is much simpler to differentiate than the original fractional form.

3. Find the second derivative, $\frac{d^2y}{dx^2}$.

   We will now differentiate the equation $(1 - x^2) \left(\frac{dy}{dx}\right)^2 = 1$ with respect to $x$. We need to apply the product rule on the left side and the chain rule for $\left(\frac{dy}{dx}\right)^2$.

   Let $u = (1 - x^2)$ and $v = \left(\frac{dy}{dx}\right)^2$.

   Then $\frac{du}{dx} = -2x$.

   And $\frac{dv}{dx} = 2\left(\frac{dy}{dx}\right) \frac{d}{dx}\left(\frac{dy}{dx}\right) = 2\frac{dy}{dx} \frac{d^2y}{dx^2}$.

   Applying the product rule, $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$: …