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Find the derivative of x^3 * sin(x)

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Answer
3x2sin⁡(x)+x3cos⁡(x)\boxed{3x^2 \sin(x) + x^3 \cos(x)}

Evaluating at x=0x = 0 gives f′(0)=0f'(0) = 0, which matches the flat slope of x3sin⁡(x)x^3 \sin(x) at the origin.

Problem: Find the derivative of f(x)=x3sin⁡(x)f(x) = x^3 \sin(x) with respect to xx.

Steps

  1. Apply the product rule (u⋅v)′=u′v+uv′(u \cdot v)' = u'v + uv' where u=x3u = x^3 and v=sin⁡(x)v = \sin(x)
    ddx[x3sin⁡(x)]=ddx[x3]⋅sin⁡(x)+x3⋅ddx[sin⁡(x)]\frac{d}{dx}[x^3 \sin(x)] = \frac{d}{dx}[x^3] \cdot \sin(x) + x^3 \cdot \frac{d}{dx}[\sin(x)]
  2. Differentiate each factor using the power rule on x3x^3 and the standard derivative of sin⁡(x)\sin(x)
    =3x2sin⁡(x)+x3cos⁡(x)= 3x^2 \sin(x) + x^3 \cos(x)
  3. Factor out the common term x2x^2
    =x2(3sin⁡(x)+xcos⁡(x))= x^2(3\sin(x) + x\cos(x))

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