Q.Find the derivative of $\dfrac{x}{\sin^n x}$.

Concept: Quotient Rule Differentiation

We differentiate $y = \frac{x}{\sin^n x}$ using the quotient rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.

Let $u = x$, so $u' = 1$.

Let $v = \sin^n x = (\sin x)^n$, so $v' = n \sin^{n-1} x \cdot \cos x$ (chain rule).

Applying the quotient rule:

Factor $\sin^{n-1} x$ from the numerator: …

The Quotient Rule is the natural choice here because we have one function divided by another. But the denominator isn't just $\sin x$ — it's $\sin^n x$, which is a power of a sine function. That means we'll need the Chain Rule inside the Quotient Rule. Let's walk through it carefully.

1. Set up the Quotient Rule.

   If $f(x) = \frac{u}{v}$, then $f'(x) = \frac{u'v - uv'}{v^2}$.

   Here, $u = x$ and $v = \sin^n x$.

2. Differentiate the numerator.

   $u' = \frac{d}{dx}(x) = 1$. That's straightforward.

3. Differentiate the denominator — this is the tricky part.

   $v = \sin^n x$ means $(\sin x)^n$. To differentiate this, use the Chain Rule:

   • The outer function is $(\text{stuff})^n$, whose derivative is $n(\text{stuff})^{n-1}$.
   • The inner function is $\sin x$, whose derivative is $\cos x$. So $v' = n (\sin x)^{n-1} \cdot \cos x = n \sin^{n-1} x \cos x$.
   Watch out

   A common mistake is to forget the Chain Rule here and write $v' = n \sin^{n-1} x$ (missing the $\cos x$). Always remember: the derivative of $\sin^n x$ is $n \sin^{n-1} x \cos x$, not just $n \sin^{n-1} x$.

4. Plug into the Quotient Rule formula.

5. Simplify the denominator. $(\sin^n x)^2 = \sin^{2n} x$. So we have:

6. Factor out the common $\sin^{n-1} x$ from the numerator. Notice $\sin^n x = \sin^{n-1} x \cdot \sin x$. So: …