Skip to content
Miscellaneous Exercise · Q14

Q.Find the derivative of sin⁡(x+a)\sin(x + a).

Chhattisgarh CgbseTextbookSubjective· 2mImportance★★★★★est
45% · 79/175 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⁡(x+a)\sin(x + a) is cos⁡(x+a)\cos(x + a). This follows directly from the chain rule — the derivative of sin⁡(u)\sin(u) is cos⁡(u)\cos(u), multiplied by the derivative of u=x+au = x + a, which is 11.

The key insight here is that sin⁡(x+a)\sin(x + a) is not a simple sine function — it's a composition of functions. The outer function is sin⁡(u)\sin(u), and the inner function is u=x+au = x + a. Whenever you have a function inside another function, the chain rule is your tool.

Let's walk through this carefully.

  1. Identify the structure.

    We have f(x)=sin⁡(x+a)f(x) = \sin(x + a). Think of it as f(x)=sin⁡(g(x))f(x) = \sin(g(x)) where g(x)=x+ag(x) = x + a. The constant aa just shifts the sine wave horizontally — it doesn't change the shape, and importantly, its derivative is zero.

  2. Recall the chain rule.

    If f(x)=sin⁡(g(x))f(x) = \sin(g(x)), then

f′(x)=cos⁡(g(x))⋅g′(x)f'(x) = \cos(g(x)) \cdot g'(x)

This is because the derivative of sin⁡(u)\sin(u) is cos⁡(u)\cos(u), and then we multiply by the derivative of the inside function u=g(x)u = g(x).

  1. Find g′(x)g'(x). Here g(x)=x+ag(x) = x + a. The derivative of xx is 11, and the derivative of a constant aa is 00. So

g′(x)=1g'(x) = 1

  1. Apply the chain rule. Put it together: f′(x)=cos⁡(x+a)⋅1=cos⁡(x+a)f'(x) = \cos(x + a) \cdot 1 = \cos(x + a) …

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.