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)…

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))

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))

Answer

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)