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

Answer 1

William Abdalla

#dy/dx=2e^xcosx#

differentiate using the #color(blue)"product rule"#

#"Given "y=g(x).h(x)" then"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=g(x)h'(x)+h(x)g'(x))color(white)(2/2)|)))#

#"here "g(x)=e^xrArrg'(x)=e^x#

#"and "h(x)=sinx+cosxrArrg'(x)=cosx-sinx#

#rArrdy/dx=e^x(cosx-sinx)+e^x(sinx+cosx)#

#color(white)(xxxxx)=e^xcosxcancel(-e^xsinx)cancel(e^xsinx)+e^xcosx#

#color(white)(xxxxx)=2e^xcosx#

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Answer 2

William Abdalla

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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Answer from HIX Tutor

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.