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Miscellaneous Exercise · Q22

Q.Find the derivative of x4(5sin⁡x−3cos⁡x)x^4(5\sin x - 3\cos x).

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We differentiate a product of two functions — x4x^4 and (5sin⁡x−3cos⁡x)(5\sin x - 3\cos x) — using the Product Rule, then simplify to get x3(20sin⁡x−12cos⁡x+5xcos⁡x+3xsin⁡x)x^3(20\sin x - 12\cos x + 5x\cos x + 3x\sin x).

When you see a product of two distinct functions, the Product Rule is your tool. The idea is simple: when both factors change, the total rate of change comes from each factor changing while the other stays fixed. If u(x)u(x) and v(x)v(x) are differentiable, then

ddx[u(x)⋅v(x)]=u′(x)⋅v(x)+u(x)⋅v′(x).\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x).

Here we have u(x)=x4u(x) = x^4 and v(x)=5sin⁡x−3cos⁡xv(x) = 5\sin x - 3\cos x. We'll differentiate each separately, then combine.


Step-by-step differentiation

  1. Identify the two factors.

    Let u=x4u = x^4 and v=5sin⁡x−3cos⁡xv = 5\sin x - 3\cos x.

  2. Differentiate the first factor.

    Using the power rule,

u′=ddx(x4)=4x3.u' = \frac{d}{dx}(x^4) = 4x^3.

  1. Differentiate the second factor. The derivative of sin⁡x\sin x is cos⁡x\cos x and the derivative of cos⁡x\cos x is −sin⁡x-\sin x, so

v′=ddx(5sin⁡x−3cos⁡x)=5cos⁡x−3(−sin⁡x)=5cos⁡x+3sin⁡x.v' = \frac{d}{dx}(5\sin x - 3\cos x) = 5\cos x - 3(-\sin x) = 5\cos x + 3\sin x.

  1. Apply the Product Rule.

ddx[x4(5sin⁡x−3cos⁡x)]=u′⋅v+u⋅v′\frac{d}{dx}[x^4(5\sin x - 3\cos x)] = u' \cdot v + u \cdot v'

=4x3⋅(5sin⁡x−3cos⁡x)+x4⋅(5cos⁡x+3sin⁡x).= 4x^3 \cdot (5\sin x - 3\cos x) + x^4 \cdot (5\cos x + 3\sin x).

  1. Expand both terms. First term: 4x3(5sin⁡x−3cos⁡x)=20x3sin⁡x−12x3cos⁡x4x^3(5\sin x - 3\cos x) = 20x^3 \sin x - 12x^3 \cos x. …

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