What is taylor 10sqrt(1+x) ?

The answer to taylor 10sqrt(1+x) is 10+5x-5/4 x^2+5/8 x^3-25/64 x^4+\ldots

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taylor 10sqrt(1+x)

taylor

10 1 + x

10 + 5 x - \frac{5}{4} x^{2} + \frac{5}{8} x^{3} - \frac{25}{64} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= 10 + \frac{\frac{d}{d x} ( 10 1 + x ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 10 1 + x ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 10 1 + x ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 10 + \frac{5}{1 !} x + \frac{- \frac{5}{2}}{2 !} x^{2} + \frac{\frac{15}{4}}{3 !} x^{3} + \frac{- \frac{75}{8}}{4 !} x^{4} + \dots

Refine

= 10 + 5 x - \frac{5}{4} x^{2} + \frac{5}{8} x^{3} - \frac{25}{64} x^{4} + \dots

\int \frac{1}{x ^{2} ( 1 - x )} d x

\frac{\partial}{\partial t} (e^{- 7 t} cos (π x))

x \to \frac{π}{4} lim (\frac{sin ( x - \frac{π}{4} )}{4 x - π})

x^{^{'}} + x = 0

\frac{1}{x} \frac{d y}{d x} = \frac{1}{x ^{2} + 1}

What is taylor 10sqrt(1+x) ?
The answer to taylor 10sqrt(1+x) is 10+5x-5/4 x^2+5/8 x^3-25/64 x^4+\ldots