Skip to content
Miscellaneous Exercise · Q19

Q.Find the derivative of sin⁡nx\sin^n x.

Jharkhand JacTextbookSubjective· 3mImportance★★★★★est
48% · 84/175 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The derivative of sin⁡nx\sin^n x is found using the chain rule: treat sin⁡x\sin x as the inner function and the power as the outer function. The result is nsin⁡n−1xcos⁡xn \sin^{n-1} x \cos x.

The key here is recognising that sin⁡nx\sin^n x is a composite function — it's not just a power of xx, but a power of sin⁡x\sin x. That means we need the chain rule, not just the power rule.

Let’s break it down.

  1. Identify the outer and inner functions.

    Write y=sin⁡nxy = \sin^n x. This means y=(sin⁡x)ny = (\sin x)^n.

    The outer function is unu^n (where uu is some expression), and the inner function is u=sin⁡xu = \sin x.

  2. Apply the chain rule.

    The chain rule says:

dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Here, dydu=nun−1\frac{dy}{du} = n u^{n-1} (power rule), and dudx=cos⁡x\frac{du}{dx} = \cos x.

  1. Substitute back.

dydx=n(sin⁡x)n−1⋅cos⁡x\frac{dy}{dx} = n (\sin x)^{n-1} \cdot \cos x

Which we write neatly as:

ddxsin⁡nx=nsin⁡n−1xcos⁡x\frac{d}{dx} \sin^n x = n \sin^{n-1} x \cos x …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.