Find the first, second, third and fourth order Maclaurin polynomials of f(x) =

arctan(x). Draw the graph of f(x) and the four polynomials on the same

diagram. (Sketch by hand or use software.)

#urgent please give me this answer and help me#

The first, second, third and fourth order Maclaurin polynomials of f(x)=arctan(x) are:

So let's start by finding the first order maclaurin polynomial:

f(x)=f(0)+f'(0)x

so let's find each part of the function:

f(0)=arctan(0)

f(0)=0

now, let's find the first derivative of f(x)

f(x)=arctan(x)

This is a usual derivative so there is a rule we can use here:

[tex]f'(x)=\frac{1}{x^{2}+1}[/tex]

so now we can find f'(0)

[tex]f'(0)=\frac{1}{(0)^{2}+1}[/tex]

f'(0)=1

So we can now complete the first order Maclaurin Polynomial:

f(x)=0+1x

which simplifies to:

f(x)=x

Now let's find the second order polynomial, for which we will need to get the second derivative of the function:

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

so:

[tex]f'(x)=\frac{1}{x^{2}+1}[/tex]

we can rewrite this derivative as:

[tex]f'(x)=(x^{2}+1)^{-1}[/tex]

and use the chain rule to get:

[tex]f''(x)=-1(x^{2}+1)^{-2}(2x)[/tex]

which simplifies to:

[tex]f''(x)=-\frac{2x}{(x^{2}+1)^{2}}[/tex]

now, we can find f''(0):

[tex]f''(0)=-\frac{2(0)}{((0)^{2}+1)^{2}}[/tex]

which yields:

f''(0)=0

so now we can complete the second order Maclaurin polynomial:

[tex]f(x)=0+1x+\frac{0}{2!}x^{2}[/tex]

which simplifies to:

f(x)=x

Now let's find the third order polynomial, for which we will need to get the third derivative of the function:

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

so:

[tex]f''(x)=-\frac{2x}{(x^{2}+1)^{2}}[/tex]

In this case we can use the quotient rule to solve this:

Quotient rule: Whenever you have a function in the form , then it's derivative is:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

in this case:

p=2x

p'=2

[tex]q=(x^{2}+1)^{2}[/tex]

[tex]q'=2(x^{2}+1)(2x)[/tex]

[tex]q'=4x(x^{2}+1)[/tex]

So when using the quotient rule we get:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

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

which simplifies to:

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

[tex]f'''(x)=\frac{6x^{2}-2}{(x^{2}+1)^{3}}[/tex]

now, we can find f'''(0):

[tex]f'''(0)=\frac{6(0)^{2}-2}{((0)^{2}+1)^{3}}[/tex]

which yields:

f'''(0)=-2

so now we can complete the third order Maclaurin polynomial:

[tex]f(x)=0+1x+\frac{0}{2!}x^{2}-\frac{2}{3!}x^{3}[/tex]

which simplifies to:

[tex]f(x)=x-\frac{1}{3}x^{3}[/tex]

Now let's find the fourth order polynomial, for which we will need to get the fourth derivative of the function:

[tex]f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^{2}+\frac{f'''(0)}{3!}x^{3}+\frac{f^{(4)}(0)}{4!}x^{4}[/tex]

so:

[tex]f'''(x)=\frac{6x^{2}-2}{(x^{2}+1)^{3}}[/tex]

In this case we can use the quotient rule to solve this:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

in this case:

[tex]p=6x^{2}-2[/tex]

p'=12x

[tex]q=(x^{2}+1)^{3}[/tex]

[tex]q'=3(x^{2}+1)^{2}(2x)[/tex]

[tex]q'=6x(x^{2}+1)^{2}[/tex]

So when using the quotient rule we get:

[tex]f'(x)=\frac{p'q-pq'}{q^{2}}[/tex]

[tex]f^{4}(x)=\frac{(12x)(x^{2}+1)^{3}-(6x^{2}-2)(6x)(x^{2}+1)^{2}}{((x^{2}+1)^{3})^{2}}[/tex]

which simplifies to:

[tex]f^{4}(x)=\frac{12x^{3}+12x-6x^{3}+12x}{(x^{2}+1)^{4}}[/tex]

[tex]f^{4}(x)=\frac{6x^{3}+24x}{(x^{2}+1)^{4}}[/tex]

now, we can find f^{4}(0):

[tex]f^{4}(x)=\frac{6(0)^{3}+24(0)}{((0)^{2}+1)^{4}}[/tex]

which yields:

[tex]f^{4}(0)=0[/tex]

so now we can complete the fourth order Maclaurin polynomial:

[tex]f(x)=0+1x+\frac{0}{2!}x^{2}-\frac{2}{3!}x^{3}+\frac{0}{4!}x^{4}[/tex]

which simplifies to:

[tex]f(x)=x-\frac{1}{3}x^{3}[/tex]

you can find the graph of the four polynomials in the attached picture.

So the first, second, third and fourth order Maclaurin polynomials of f(x)=arctan(x) are:

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