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NCERT Exemplar · Q38

Q.Differentiate with respect to xx: (sin⁡x+cos⁡x)2(\sin x + \cos x)^2.

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The derivative of (sin⁡x+cos⁡x)2(\sin x + \cos x)^2 is 2cos⁡2x2\cos 2x. We simplify the square first using a trig identity, then differentiate.

Concept First: Why Simplify Before Differentiating?

When you see a function like (sin⁡x+cos⁡x)2(\sin x + \cos x)^2, your first instinct might be to reach for the chain rule. That works, but it's messy. A cleaner path: simplify the expression first using algebraic identities. The square of a sum expands, and the resulting terms often collapse into something much simpler using trigonometric identities. This isn't just a trick — it's a core strategy: always simplify before differentiating when possible. It reduces algebra errors and often reveals the derivative in a more elegant form.

The derivative of a function at a point tells you the instantaneous rate of change. Here, we want the derivative of (sin⁡x+cos⁡x)2(\sin x + \cos x)^2 with respect to xx, meaning we treat xx as the variable and apply standard differentiation rules.


Step-by-Step Solution

1. Expand the square.

We have:

(sin⁡x+cos⁡x)2=sin⁡2x+2sin⁡xcos⁡x+cos⁡2x(\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x

2. Use the Pythagorean identity.

Recall that sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. So:

sin⁡2x+cos⁡2x+2sin⁡xcos⁡x=1+2sin⁡xcos⁡x\sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x

3. Use the double-angle identity for sine.

We know 2sin⁡xcos⁡x=sin⁡2x2\sin x \cos x = \sin 2x. Therefore:

(sin⁡x+cos⁡x)2=1+sin⁡2x(\sin x + \cos x)^2 = 1 + \sin 2x

Tip

This simplification is the key insight. Instead of differentiating a square of a sum, we now have a simple sum of a constant and a sine function. The derivative of a constant is zero, so we only need to differentiate sin⁡2x\sin 2x.

4. Differentiate term by term.

Let y=1+sin⁡2xy = 1 + \sin 2x. Then: …

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