# How do you find the taylor series for #y=(e^x)cos(x)#?

Successive derivatives look like this:

#{:(f^((0))(x) = e^x cos x, f^((0))(0) = color(blue)(1)), (f^((1))(x) = e^x (cos x - sin x), f^((1))(0) = color(blue)(1)), (f^((2))(x) = -2e^x sin x, f^((2))(0) = color(blue)(0)), (f^((3))(x) = -2e^x(cos x + sin x), f^((3))(0) = color(blue)(-2)), (f^((4))(x) = -4e^x cos x, f^((4))(0) = -4), (f^((5))(x) = -4e^x(cos x - sin x), f^((5))(0) = -4), (f^((6))(x) = 8e^x sin x, f^((6))(0) = 0), (f^((7))(x) = 8e^x(cos x + sin x), f^((7))(0) = 8):}#

In our case, capture the recurring pattern as follows:

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To find the Taylor series for ( y = e^x \cos(x) ), follow these steps:

- Write down the Taylor series expansions for ( e^x ) and ( \cos(x) ).
- Multiply the two series term by term.
- Combine like terms and simplify to obtain the Taylor series for ( y = e^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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