What is maclaurin (1+x)^{-3/5} ?

The answer to maclaurin (1+x)^{-3/5} is 1-3/5 x+12/25 x^2-52/125 x^3+234/625 x^4+\ldots

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maclaurin (1+x)^{-3/5}

maclaurin

(1 + x)^{- \frac{3}{5}}

1 - \frac{3}{5} x + \frac{12}{25} x^{2} - \frac{52}{125} x^{3} + \frac{234}{625} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( ( 1 + x ) ^{- \frac{3}{5}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ( 1 + x ) ^{- \frac{3}{5}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ( 1 + x ) ^{- \frac{3}{5}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{- \frac{3}{5}}{1 !} x + \frac{\frac{24}{25}}{2 !} x^{2} + \frac{- \frac{312}{125}}{3 !} x^{3} + \frac{\frac{5616}{625}}{4 !} x^{4} + \dots

Refine

= 1 - \frac{3}{5} x + \frac{12}{25} x^{2} - \frac{52}{125} x^{3} + \frac{234}{625} x^{4} + \dots

x \to 7 lim (\frac{x + 7}{∣ x + 7 ∣})

x \to 2 lim (x^{4} - 8)

\int (sec (5 θ) tan (5 θ) + csc^{2} (2 θ)) d θ

\frac{\partial}{\partial y} (x y cos (yz))

\frac{d y}{d x} = \frac{x ^{2} + 3}{y}

What is maclaurin (1+x)^{-3/5} ?
The answer to maclaurin (1+x)^{-3/5} is 1-3/5 x+12/25 x^2-52/125 x^3+234/625 x^4+\ldots