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Find 
f^(')(x) where 
f(x)=2x cos(x).

Find f(x) f^{\prime}(x) where f(x)=2xcos(x) f(x)=2 x \cos (x) .

Full solution

Q. Find f(x) f^{\prime}(x) where f(x)=2xcos(x) f(x)=2 x \cos (x) .
  1. Identify Components: Identify the components of the function that will require the use of the product rule for differentiation.\newlineThe function f(x)=2xcos(x)f(x) = 2x \cos(x) is a product of two functions, g(x)=2xg(x) = 2x and h(x)=cos(x)h(x) = \cos(x).
  2. Recall Product Rule: Recall the product rule for differentiation, which states that if f(x)=g(x)h(x)f(x) = g(x) \cdot h(x), then f(x)=g(x)h(x)+g(x)h(x)f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x). We will apply this rule to find the derivative of f(x)=2xcos(x)f(x) = 2x \cos(x).
  3. Differentiate g(x)g(x): Differentiate g(x)=2xg(x) = 2x with respect to xx. The derivative of g(x)g(x) with respect to xx is g(x)=ddx(2x)=2g'(x) = \frac{d}{dx}(2x) = 2.
  4. Differentiate h(x)h(x): Differentiate h(x)=cos(x)h(x) = \cos(x) with respect to xx. The derivative of h(x)h(x) with respect to xx is h(x)=ddx(cos(x))=sin(x)h'(x) = \frac{d}{dx}(\cos(x)) = -\sin(x).
  5. Apply Product Rule: Apply the product rule using the derivatives found in the previous steps.\newlineUsing g(x)=2g'(x) = 2 and h(x)=sin(x)h'(x) = -\sin(x), we get f(x)=g(x)h(x)+g(x)h(x)=2cos(x)+2x(sin(x))f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x) = 2 \cdot \cos(x) + 2x \cdot (-\sin(x)).
  6. Simplify Derivative: Simplify the expression for the derivative.\newlinef(x)=2cos(x)2xsin(x)f'(x) = 2 \cdot \cos(x) - 2x \cdot \sin(x).\newlineThis is the derivative of the function f(x)=2xcos(x)f(x) = 2x \cos(x).

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