What is taylor 100x+0.01x^2 ?

The answer to taylor 100x+0.01x^2 is 100x+0.01x^2

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taylor 100x+0.01x^2

taylor

100 x + 0.01 x^{2}

100 x + 0.01 x^{2}

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( 100 x + 0.01 x ^{2} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 100 x + 0.01 x ^{2} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 100 x + 0.01 x ^{2} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{100}{1 !} x + \frac{0.02}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{0}{10 !} x^{10}

Refine

= 100 x + 0.01 x^{2}

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

\int_{0}^{l n (2)} 4 (2 - x e^{x}) d x

x^{2} \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x} + y = 0

f (x) = e^{t a n (x)}

\frac{d}{d x} (1 - (x^{2} + y^{2})^{\frac{1}{3}})