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

The answer to taylor (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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taylor (1+x)^{-3/5}

taylor

(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 Taylor 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

\frac{d}{d x} (17 e^{x^{2}})

y^{^{''}} + 8 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

(x^{4} + 2)^{2} (x^{3} + 4)^{4}

\int 9 sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

f (3) = 3 x^{2} + 15

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