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NCERT Exemplar · Q40

Q.Differentiate with respect to xx: x2sin⁡x+cos⁡2xx^2 \sin x + \cos 2x.

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Differentiate each term separately: the first needs the product rule (since x2x^2 and sin⁡x\sin x are multiplied), the second uses the chain rule. The derivative is x2cos⁡x+2xsin⁡x−2sin⁡2x\boxed{x^2 \cos x + 2x \sin x - 2 \sin 2x}.

The expression x2sin⁡x+cos⁡2xx^2 \sin x + \cos 2x is a sum of two terms, so we can differentiate each independently and add the results. The first term is a product of two functions of xx, which calls for the product rule. The second term is a composite function, requiring the chain rule.

Differentiating x2sin⁡xx^2 \sin x

When two functions are multiplied, their derivative is NOT simply the product of their derivatives. Instead, the product rule tells us:

ddx[u⋅v]=u′v+uv′\frac{d}{dx}[u \cdot v] = u' v + u v'

This captures the idea that both factors contribute to the rate of change.

1. Identify u=x2u = x^2 and v=sin⁡xv = \sin x.

2. Find the derivatives: u′=2xu' = 2x and v′=cos⁡xv' = \cos x.

3. Apply the product rule:

ddx(x2sin⁡x)=(2x)(sin⁡x)+(x2)(cos⁡x)=2xsin⁡x+x2cos⁡x\frac{d}{dx}(x^2 \sin x) = (2x)(\sin x) + (x^2)(\cos x) = 2x \sin x + x^2 \cos x

Tip

In the product rule, the order doesn't matter for the final sum, but keeping a consistent pattern (first function differentiated × second unchanged, then first unchanged × second differentiated) helps avoid sign errors.

Differentiating cos⁡2x\cos 2x

The function cos⁡2x\cos 2x is a composition: the outer function is cosine, the inner function is 2x2x. …

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