Animated Solution for Mathematics - Differentiation: If ${\cos }^{-1}\left(\frac{y}{2}\right)={\log }_e{\left(\frac{x}{5}\right)}^5,|y|<2$, then :

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of inverse trigonometry and logarithms.

But here is the secret of JEE Advanced: the most intimidating problems are often just simple problems wearing a disguise. Our goal is to strip away that disguise.

The Logarithmic Clean-Up

We start with the equation:

Your instinct might be to start differentiating immediately. Stop. Take a breath.

Look at that power of $5$ inside the logarithm. In the world of calculus, exponents inside logs are gifts. We use the property ${\log }_e(a^b)=b{\log }_e(a)$ to bring that $5$ down.

Suddenly, the equation becomes:

See how much lighter that feels? We have transformed a complex power into a simple coefficient.

The First Derivative Dance

Now, we apply the derivative operator $\frac{d}{dx}$ to both sides. On the left, we have the derivative of ${\cos }^{-1}(u)$.

Recall that:

Here, $u=\frac{y}{2}$, so $\frac{du}{dx}=\frac{y^{\prime }}{2}$. This gives us:

On the right, the derivative of $5{\log }_e(\frac{x}{5})$ is simply $5\cdot \frac{1}{x/5}\cdot \frac{1}{5}$, which simplifies beautifully to $\frac{5}{x}$. Equating these, we get:

The Strategic Pivot

Look at the denominator on the left: $2\sqrt{1-\frac{y^2}{4}}$. If we simplify the term inside the square root, we get $\sqrt{\frac{4-y^2}{4}}$, which is $\frac{\sqrt{4-y^2}}{2}$.

The $2$ in the denominator cancels perfectly with the $2$ outside! We are left with:

Now, we face a choice: differentiate again with a quotient rule and a radical, or be clever. Let's be clever.

We cross-multiply to get $-x{y}^{\prime }=5\sqrt{4-y^2}$. Then, we square both sides:

This is the turning point. We have eliminated the radical, and the path to the second derivative is now clear.

The Final Product Rule

We differentiate $x^2(y^{\prime }{)}^2=25(4-y^2)$ with respect to $x$. On the left, we use the product rule: $\frac{d}{dx}[x^2]\cdot (y^{\prime }{)}^2+x^2\cdot \frac{d}{dx}[(y^{\prime }{)}^2]$.

This gives us:

Notice that every term contains a $2y^{\prime }$. Dividing by $2y^{\prime }$ (assuming $y' eq 0$), we get:

Rearranging this gives us the final, elegant result:

Look at that. We started with a complex logarithmic-trigonometric equation and arrived at a clean, second-order differential equation.

The key wasn't brute force; it was simplification, strategic squaring, and careful application of the chain and product rules. You have the tools. Trust your process, and keep solving.