What is taylor x^2,1 ?

The answer to taylor x^2,1 is 1+2(x-1)+(x-1)^2

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taylor x^2,1

taylor

x^{2}, 1

1 + 2 (x - 1) + (x - 1)^{2}

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{2}{1 !} (x - 1) + \frac{2}{2 !} (x - 1)^{2} + \frac{0}{3 !} (x - 1)^{3} + \frac{0}{4 !} (x - 1)^{4} + \frac{0}{5 !} (x - 1)^{5} + \frac{0}{6 !} (x - 1)^{6} + \frac{0}{7 !} (x - 1)^{7} + \frac{0}{8 !} (x - 1)^{8} + \frac{0}{9 !} (x - 1)^{9} + \frac{0}{10 !} (x - 1)^{10}

Refine

= 1 + 2 (x - 1) + (x - 1)^{2}

y^{^{''''}} + y^{^{''}} = 0

\frac{\partial}{\partial x} (x z - e^{x} sin (y))

x \to - 2 lim (\frac{5 x ^{2} + 6}{x})

(\frac{1}{( 1 - x ) ^{2}})^{^{'}}

y = \frac{1}{4 x ^{2}}, y = x, y = \frac{x}{10}