Question

question 9 of 11 if (f(x)=e^{x}(sin x+cos x)), then (f(x)=) options: (e^{x}cos x), (2e^{x}cos x), (e^{x}(cos^{2}x - sin^{2}x)), (2e^{x}(cos x+sin x))

Explanation:

Step1: Apply product - rule

The product - rule states that if \(y = u\cdot v\), then \(y'=u'v + uv'\). Here, \(u = e^{x}\) and \(v=\sin x+\cos x\). The derivative of \(u = e^{x}\) is \(u'=e^{x}\), and the derivative of \(v=\sin x+\cos x\) is \(v'=\cos x-\sin x\).

Step2: Calculate \(f'(x)\)

\[

$$\begin{align*} f'(x)&=e^{x}(\sin x+\cos x)+e^{x}(\cos x - \sin x)\\ &=e^{x}\sin x+e^{x}\cos x+e^{x}\cos x - e^{x}\sin x\\ &=2e^{x}\cos x \end{align*}$$

\]

Answer:

\(2e^{x}\cos x\)