Q.Differentiate with respect to $x$: $\sin^3 x \cos^3 x$.

The key idea is to use the Quotient Rule after rewriting the product as a single trigonometric function, or to differentiate directly using the Product Rule and Chain Rule. Here, the product rule is more direct.

Let $y = \sin^3 x \cos^3 x$. Write it as $y = (\sin x)^3 (\cos x)^3$.

Differentiate using the product rule:

$y' = 3\sin^2 x \cos x \cdot \cos^3 x + \sin^3 x \cdot 3\cos^2 x (-\sin x)$.

Factor common terms:

$y' = 3\sin^2 x \cos^2 x (\cos^2 x - \sin^2 x)$. …

The function we have is a product of two functions, each of which is a power of a trigonometric function. The most direct way is to apply the product rule: if $y = u \cdot v$, then $\frac{dy}{dx} = u'v + uv'$. Here, $u = \sin^3 x$ and $v = \cos^3 x$.

But each of these is itself a composite function. For example, $\sin^3 x$ means $(\sin x)^3$ — the outer function is "cube", the inner is $\sin x$. So to differentiate $u$ and $v$, we need the chain rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$.

Let’s work through it step by step.

1. Differentiate $u = \sin^3 x$.

   The outer function is $(\cdot)^3$, derivative $3(\cdot)^2$. The inner function is $\sin x$, derivative $\cos x$.

   So $u' = 3(\sin x)^2 \cdot \cos x = 3\sin^2 x \cos x$.

2. Differentiate $v = \cos^3 x$.

   Outer: $(\cdot)^3$, derivative $3(\cdot)^2$. Inner: $\cos x$, derivative $-\sin x$.

   So $v' = 3(\cos x)^2 \cdot (-\sin x) = -3\cos^2 x \sin x$.

3. Apply the product rule.

4. Simplify each term.

   First term: $3\sin^2 x \cos x \cdot \cos^3 x = 3\sin^2 x \cos^4 x$.

   Second term: $-3\sin^3 x \cdot \cos^2 x \sin x = -3\sin^4 x \cos^2 x$.

   So we have:

5. Factor common terms. Both terms have $3\sin^2 x \cos^2 x$ in common. Factor it out:

6. Recognize the trigonometric identities.

   • $\sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x$ (since $\sin 2x = 2\sin x \cos x$, so $\sin^2 2x = 4\sin^2 x \cos^2 x$).
   • $\cos^2 x - \sin^2 x = \cos 2x$.

   So the derivative can also be written as:

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