What is taylor sqrt(8+x^2),1 ?

The answer to taylor sqrt(8+x^2),1 is 3+1/3 (x-1)+4/27 (x-1)^2-4/243 (x-1)^3-4/2187 (x-1)^4+\ldots

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taylor sqrt(8+x^2),1

taylor

8 + x^{2}, 1

3 + \frac{1}{3} (x - 1) + \frac{4}{27} (x - 1)^{2} - \frac{4}{243} (x - 1)^{3} - \frac{4}{2187} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

= 3 + \frac{\frac{d}{d x} ( 8 + x ^{2} ) ( 1 )}{1 !} (x - 1) + \frac{\frac{d ^{2}}{d x ^{2}} ( 8 + x ^{2} ) ( 1 )}{2 !} (x - 1)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 8 + x ^{2} ) ( 1 )}{3 !} (x - 1)^{3} + \dots

Evaluate Derivatives

= 3 + \frac{\frac{1}{3}}{1 !} (x - 1) + \frac{\frac{8}{27}}{2 !} (x - 1)^{2} + \frac{- \frac{8}{81}}{3 !} (x - 1)^{3} + \frac{- \frac{32}{729}}{4 !} (x - 1)^{4} + \dots

Refine

= 3 + \frac{1}{3} (x - 1) + \frac{4}{27} (x - 1)^{2} - \frac{4}{243} (x - 1)^{3} - \frac{4}{2187} (x - 1)^{4} + \dots

f (x) = 4 x^{5} ln (x)

ln (1 - x^{2})

\int t^{2} e^{\frac{t}{3}} d t

x \to 4 lim (\frac{5}{x - 4})

x \to \infty lim (\frac{( e ^{- x y - x + y} )}{x ^{2} y ^{2} + x y})

What is taylor sqrt(8+x^2),1 ?
The answer to taylor sqrt(8+x^2),1 is 3+1/3 (x-1)+4/27 (x-1)^2-4/243 (x-1)^3-4/2187 (x-1)^4+\ldots