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{:[f(x)=8e^(x cos x)],[f^(')(x)=]:}

f(x)=8excos⁡xf′(x)= \begin{array}{r}f(x)=8 e^{x \cos x} \\ f^{\prime}(x)=\end{array}

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Full solution

Q. f(x)=8excos⁡xf′(x)= \begin{array}{r}f(x)=8 e^{x \cos x} \\ f^{\prime}(x)=\end{array}
  1. Identify function and type: Identify the function and the type of differentiation required. \newlinef(x)=8excos⁡xf(x) = 8e^{x \cos x}, need to find f′(x)f'(x) using the chain rule and product rule.
  2. Apply chain and product rule: Apply the chain rule and product rule.\newlineLet u=xcos⁡xu = x \cos x, then f(x)=8euf(x) = 8e^u.\newlineUsing the chain rule, f′(x)=8eu⋅dudxf'(x) = 8e^u \cdot \frac{du}{dx}.\newlineNow, find dudx\frac{du}{dx} using the product rule: dudx=ddx(xcos⁡x)=cos⁡x+x(−sin⁡x)=cos⁡x−xsin⁡x\frac{du}{dx} = \frac{d}{dx}(x \cos x) = \cos x + x(-\sin x) = \cos x - x \sin x.
  3. Substitute to find f′(x)f'(x): Substitute back to find f′(x)f'(x).\newlinef′(x)=8e(xcos⁡x)⋅(cos⁡x−xsin⁡x)f'(x) = 8e^{(x \cos x)} \cdot (\cos x - x \sin x).

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