Animated Solution for Mathematics - Differentiation: If ${\log }_{e}y=3{\sin }^{-1}x$, then $(1-x^2)y^{\prime \prime }-x{y}^{\prime }$ at $x=\frac{1}{2}$ is equal to

My dear student, welcome to a problem that is less about brute force and more about the art of mathematical strategy. When you first look at the expression $(1-x^2)y^{\prime \prime }-x{y}^{\prime }$, it might seem like a daunting task of repeated differentiation.

But I want you to pause. In the world of JEE Advanced, whenever you see a second-order derivative expression like this, it is rarely a call to just differentiate blindly. It is a call to find a hidden relationship.

We begin with the given equation: ${\log }_{e}y=3{\sin }^{-1}x$. While we could differentiate this implicitly, it is far more elegant to work with the explicit form.

By the definition of logarithms, we can rewrite this as:

This is our starting point. It is clean, it is explicit, and it is ready for action.

Now, we differentiate with respect to $x$. Using the Chain Rule, the derivative of $e^{3{\sin }^{-1}x}$ is $e^{3{\sin }^{-1}x}$ multiplied by the derivative of the exponent, which is $3\cdot \frac{1}{\sqrt{1-x^2}}$.

So, we have:

Notice how $y$ has reappeared! This is the beauty of exponential functions. Instead of carrying that bulky exponential term, we simply substitute $y$ back in. This gives us:

Here is where the magic happens. We could use the quotient rule, but why invite trouble? Let us cross-multiply to get $y^{\prime }\sqrt{1-x^2}=3y$.

Now, we have a clean, linear-looking equation. To make the next differentiation effortless, we square both sides to eliminate the radical:

This is the turning point. We have transformed a radical equation into a polynomial one.

Now, we differentiate this squared equation with respect to $x$. Applying the product rule on the left side, the derivative of $(y^{\prime }{)}^2$ is $2y^{\prime }{y}^{\prime \prime }$, and the derivative of $(1-x^2)$ is $-2x$.

So, we get:

Look closely at this equation. Every single term contains $2y^{\prime }$. Since $y^{\prime }$ is not zero, we can divide the entire equation by $2y^{\prime }$. What remains is pure gold:

We have arrived at our destination. The expression we were asked to evaluate is exactly $9y$. All that remains is to find the value of $y$ at $x=\frac{1}{2}$.

Substituting $x=\frac{1}{2}$ into our original equation, we get:

Since ${\sin }^{-1}(1/2)=\frac{\pi }{6}$, the exponent becomes $3\cdot \frac{\pi }{6}=\frac{\pi }{2}$. Thus, $y=e^{\pi /2}$.

Our final answer is $9e^{\pi /2}$. See how we avoided the chaos of the quotient rule and arrived at the answer with grace? This, my friend, is the essence of JEE Advanced mathematics.