What is taylor 9sqrt(1+x) ?

The answer to taylor 9sqrt(1+x) is 9+9/2 x-9/8 x^2+9/16 x^3-45/128 x^4+\ldots

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taylor 9sqrt(1+x)

taylor

9 1 + x

9 + \frac{9}{2} x - \frac{9}{8} x^{2} + \frac{9}{16} x^{3} - \frac{45}{128} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= 9 + \frac{\frac{d}{d x} ( 9 1 + x ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 9 1 + x ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 9 1 + x ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 9 + \frac{\frac{9}{2}}{1 !} x + \frac{- \frac{9}{4}}{2 !} x^{2} + \frac{\frac{27}{8}}{3 !} x^{3} + \frac{- \frac{135}{16}}{4 !} x^{4} + \dots

Refine

= 9 + \frac{9}{2} x - \frac{9}{8} x^{2} + \frac{9}{16} x^{3} - \frac{45}{128} x^{4} + \dots

sin^{3} (4 x)

\int 3 x^{2} (x^{3} - 4)^{4} d x

\frac{d}{d x} (\frac{2 x}{e ^{4 x}})

y^{^{'}} = x + y

y^{^{''}} + y = sin (6 t), y (0) = 0, y^{^{'}} (0) = 0

What is taylor 9sqrt(1+x) ?
The answer to taylor 9sqrt(1+x) is 9+9/2 x-9/8 x^2+9/16 x^3-45/128 x^4+\ldots