Animated Solution for Mathematics - Differentiation: If $y={\tan }^{-1}(\sec x^3-\tan x^3)$, $\frac{\pi }{2}<x^3<\frac{3\pi }{2}$, then

Peeling Back the Layers

Welcome, fellow traveler on the JEE journey. Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of calculus.

We are given $y={\tan }^{-1}(\sec x^3-\tan x^3)$ with a specific domain constraint. Many students would immediately reach for the chain rule, but that is a path to exhaustion.

In the world of JEE Advanced, the secret is often not in the brute force, but in the elegance of simplification. Let us peel back the layers together.

The Imposters

Look at the expression inside the inverse tangent: $\sec x^3-\tan x^3$. These are what I call 'imposters.' They hide the true nature of the function.

To see the reality, we must strip them down to their core: sine and cosine. We know that $\sec x^3=\frac{1}{\cos x^3}$ and $\tan x^3=\frac{\sin x^3}{\cos x^3}$.

By substituting these, our expression becomes:

Suddenly, the fog begins to lift. We have a single fraction, a much more manageable beast.

The Complementary Angle Trick

Now, we want to use half-angle identities, but they work best with $1-\cos A$, not $1-\sin A$. How do we bridge this gap?

We use the beauty of complementary angles. We know that $\sin \theta =\cos (\frac{\pi }{2}-\theta )$ and $\cos \theta =\sin (\frac{\pi }{2}-\theta )$.

By applying this to our expression, the numerator becomes $1-\cos (\frac{\pi }{2}-x^3)$ and the denominator becomes $\sin (\frac{\pi }{2}-x^3)$. This is the 'Aha!' moment.

Half-Angle Magic

Let $A=\frac{\pi }{2}-x^3$. Our expression is now:

Using the identities $1-\cos A=2{\sin }^2(\frac{A}{2})$ and $\sin A=2\sin (\frac{A}{2})\cos (\frac{A}{2})$, the expression simplifies to:

Substituting $A$ back, we get $\tan (\frac{\pi }{4}-\frac{x^3}{2})$.

The Domain Trap

We are now at $y={\tan }^{-1}(\tan (\frac{\pi }{4}-\frac{x^3}{2}))$. Can we just cancel the ${\tan }^{-1}$ and $\tan$? Only if the angle is in the principal domain $(-\frac{\pi }{2},\frac{\pi }{2})$.

Let us check. We are given $\frac{\pi }{2}<x^3<\frac{3\pi }{2}$. Dividing by $-2$ gives $-\frac{3\pi }{4}<-\frac{x^3}{2}<-\frac{\pi }{4}$.

Adding $\frac{\pi }{4}$ gives $-\frac{\pi }{2}<\frac{\pi }{4}-\frac{x^3}{2}<0$. Since this interval is entirely within $(-\frac{\pi }{2},\frac{\pi }{2})$, we are safe! The function simplifies to:

The Calculus Finale

Now, the calculus becomes trivial. Differentiating $y=\frac{\pi }{4}-\frac{x^3}{2}$ with respect to $x$ gives $y^{\prime }=-\frac{3x^2}{2}$.

Differentiating again gives $y^{\prime \prime }=-3x$. We need to match this to the options. Multiplying by $x^2$, we get $x^2{y}^{\prime \prime }=-3x^3$.

From our simplified $y$, we know $x^3=\frac{\pi }{2}-2y$. Substituting this in:

Rearranging, we arrive at the final result:

We have conquered the problem!