4th degree maclaurin polynomial of -1/((1+x)^2)

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Find 0th Derivative: To find the Maclaurin polynomial of a function, we need to calculate the derivatives of the function at x = 0 up to the degree we are interested in, which is the 4th degree in this case.Find 1st Derivative: First, let's find the 0th derivative, which is the function itself. For -1/((1+x)^2), when x = 0, the value is -1/(1+0)^2 = -1.Find 2nd Derivative: Now, let's find the 1st derivative of -1/((1+x)^2). The derivative is 2/(1+x)^3. When x = 0, the value is 2/(1+0)^3 = 2.Find 3rd Derivative: Next, we find the 2nd derivative. The 2nd derivative of 2/(1+x)^3 is -6/(1+x)^4. When x = 0, the value is -6/(1+0)^4 = -6.Find 4th Derivative: We continue with the 3rd derivative. The 3rd derivative of -6/(1+x)^4 is 24/(1+x)^5. When x = 0, the value is 24/(1+0)^5 = 24.Construct Maclaurin Polynomial: Finally, we find the 4th derivative. The 4th derivative of 24/(1+x)^5 is -120/(1+x)^6. When x = 0, the value is -120/(1+0)^6 = -120.Substitute Values: Now we have all the derivatives at x = 0. We can construct the Maclaurin polynomial using the formula: P(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + f''''(0)x^4/4!Simplify Polynomial: Substituting the values we found into the Maclaurin polynomial formula, we get: P(x) = -1 + 2x - 6x^2/2! + 24x^3/3! - 120x^4/4!Simplify the polynomial by calculating the factorials: P(x) = -1 + 2x - 3x^2 + 4x^3 - 5x^4

$4$ th degree maclaurin polynomial of $-\frac{1}{(1+x)^2}$

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$4$ th degree maclaurin polynomial of $-\frac{1}{(1+x)^2}$