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Find f′(x)f'(x) where f(x)=−2xcos⁡(x)f(x)=-2x\cos(x)

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Full solution

Q. Find f′(x)f'(x) where f(x)=−2xcos⁡(x)f(x)=-2x\cos(x)
  1. Apply Product Rule: Apply the product rule to find the derivative of f(x)=−2xcos⁡(x)f(x)=-2x \cos(x).\newlineThe product rule states that if you have a function that is the product of two functions, u(x)u(x) and v(x)v(x), then the derivative of this function is given by u′(x)v(x)+u(x)v′(x)u'(x)v(x) + u(x)v'(x). Here, u(x)=−2xu(x) = -2x and v(x)=cos⁡(x)v(x) = \cos(x).
  2. Differentiate −2x-2x: Differentiate u(x)=−2xu(x) = -2x with respect to xx. The derivative of −2x-2x with respect to xx is −2-2, since the derivative of xx is 11.
  3. Differentiate cos⁡(x)\cos(x): Differentiate v(x)=cos⁡(x)v(x) = \cos(x) with respect to xx. The derivative of cos⁡(x)\cos(x) with respect to xx is −sin⁡(x)-\sin(x).
  4. Apply Product Rule: Apply the product rule using the derivatives from Step 22 and Step 33.\newlineUsing the product rule, we have:\newlinef′(x)=u′(x)v(x)+u(x)v′(x)f'(x) = u'(x)v(x) + u(x)v'(x)\newlinef′(x)=(−2)⋅cos⁡(x)+(−2x)⋅(−sin⁡(x))f'(x) = (-2) \cdot \cos(x) + (-2x) \cdot (-\sin(x))
  5. Simplify Derivative: Simplify the expression for the derivative. f′(x)=−2cos⁡(x)+2xsin⁡(x)f'(x) = -2\cos(x) + 2x\sin(x)

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