# How do you find the Maclaurin series of #f(x)=ln(1+x^2)# ?

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To find the Maclaurin series of ( f(x) = \ln(1 + x^2) ), we'll first find the derivatives of ( f(x) ) and then evaluate them at ( x = 0 ) to obtain the coefficients of the Maclaurin series. The Maclaurin series expansion is given by:

[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots ]

First, let's find the derivatives of ( f(x) = \ln(1 + x^2) ):

[ f'(x) = \frac{2x}{1 + x^2} ] [ f''(x) = \frac{2(1 + x^2) - 2x(2x)}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} ] [ f'''(x) = \frac{(-4x)(1 + x^2)^2 - 2(2 - 2x^2)(2x)(1 + x^2)}{(1 + x^2)^4} = \frac{-4x - 8x^3 + 8x^3}{(1 + x^2)^3} = \frac{-4x}{(1 + x^2)^3} ]

Now, let's evaluate these derivatives at ( x = 0 ) to find the coefficients of the Maclaurin series:

[ f(0) = \ln(1) = 0 ] [ f'(0) = \frac{2 \cdot 0}{1 + 0^2} = 0 ] [ f''(0) = \frac{2 - 2 \cdot 0^2}{(1 + 0^2)^2} = 2 ] [ f'''(0) = \frac{-4 \cdot 0}{(1 + 0^2)^3} = 0 ]

Therefore, the Maclaurin series of ( f(x) = \ln(1 + x^2) ) is:

[ f(x) = 0 + 0x + \frac{2}{2!}x^2 + 0x^3 + \cdots ]

[ f(x) = x^2 ]

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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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