# How do you find the derivative of #y=e^x(sinx+cosx)#?

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To find the derivative of ( y = e^x (\sin x + \cos x) ), you can apply the product rule and chain rule.

Using the product rule:

[ \frac{d}{dx} (f(x)g(x)) = f'(x)g(x) + f(x)g'(x) ]

Let ( f(x) = e^x ) and ( g(x) = \sin x + \cos x ). Then:

[ f'(x) = e^x \quad \text{and} \quad g'(x) = \cos x - \sin x ]

Applying the product rule:

[ \frac{d}{dx} (e^x (\sin x + \cos x)) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) ]

Simplify:

[ \frac{d}{dx} (e^x (\sin x + \cos x)) = e^x \sin x + e^x \cos x + e^x \cos x - e^x \sin x ]

Combining like terms:

[ \frac{d}{dx} (e^x (\sin x + \cos x)) = \boxed{2e^x \cos x} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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