Animated Solution for Mathematics - Differentiation: If $y=\frac{(\sqrt{x}+1)(x^2-\sqrt{x})}{x\sqrt{x}+x+\sqrt{x}}+\frac{1}{15}(3{\cos }^{2}x-5){\cos }^{3}x$, then $96y^{\prime }(\frac{\pi }{6})$ is equal to :

My dear student, take a deep breath. When you look at an expression like

it is natural to feel a surge of intimidation. It looks like a monster, doesn't it?

But here is the secret of the JEE Advanced: the examiners do not want to test your ability to perform tedious, soul-crushing calculations. They want to test your ability to see the elegance hidden beneath the chaos. This problem is not a test of your stamina; it is a test of your vision.

The Algebraic Surgery

Let us perform some surgery on that first term. If you try to differentiate that fraction directly, you will be lost in a forest of quotient rules.

Instead, let us look at the numerator: $x^2-\sqrt{x}$. If we factor out $\sqrt{x}$, we get $\sqrt{x}(x^{3/2}-1)$.

Now, recognize that $x^{3/2}$ is just $(\sqrt{x}{)}^3$. This is a classic $a^3-b^3$ structure! Using the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$, we can rewrite the numerator as:

Now, look at the denominator: $x\sqrt{x}+x+\sqrt{x}$. If we factor out $\sqrt{x}$ here, we get $\sqrt{x}(x+\sqrt{x}+1)$.

Do you see it? The term $(x+\sqrt{x}+1)$ appears in both the numerator and the denominator. They cancel out beautifully, leaving us with just $(\sqrt{x}+1)(\sqrt{x}-1)$, which is simply $x-1$.

That entire, terrifying fraction has collapsed into a simple linear expression!

The Trigonometric Cleanup

Now, let us turn our attention to the second term: $\frac{1}{15}(3{\cos }^{2}x-5){\cos }^{3}x$. Again, do not reach for the product rule.

Distribute the ${\cos }^{3}x$ and the constant $\frac{1}{15}$ inside the bracket. This gives us:

Our function $y$ is now a clean, manageable expression:

The Calculus

Now that we have simplified the function, differentiation becomes a joy. The derivative of $x-1$ is simply $1$.

For the trigonometric parts, we use the chain rule. The derivative of $\frac{1}{5}{\cos }^{5}x$ is:

Similarly, the derivative of $-\frac{1}{3}{\cos }^{3}x$ is:

Assembling these, we get:

The Final Evaluation

We are at the finish line. We need to evaluate $y^{\prime }(\frac{\pi }{6})$. We know that $\sin (\frac{\pi }{6})=\frac{1}{2}$ and $\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}$.

Substituting these values, we get:

Calculating the powers, we have:

Converting to a common denominator of $32$, we get:

Finally, the question asks for $96y^{\prime }(\frac{\pi }{6})$, which is:

Isn't it marvelous? A problem that started as a nightmare ended with such a clean, integer result. This is the beauty of mathematics—if you approach it with patience and strategy, the complexity always gives way to elegance.