What is maclaurin e^{2x^3} ?

The answer to maclaurin e^{2x^3} is 1+2x^3+2x^6+4/3 x^9+\ldots

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maclaurin e^{2x^3}

maclaurin

e^{2 x^{3}}

1 + 2 x^{3} + 2 x^{6} + \frac{4}{3} x^{9} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{0}{2 !} x^{2} + \frac{12}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{1440}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{483840}{9 !} x^{9} + \frac{0}{10 !} x^{10} + \dots

Refine

= 1 + 2 x^{3} + 2 x^{6} + \frac{4}{3} x^{9} + \dots

\frac{\partial}{\partial y} (4 x^{3} y^{2} - 1)

(y p + 2 y) (p d y + y d p) - (2 y p + 4 y) d y = 0

cos^{2} (13 t)

y = ln (5)

\frac{d y}{d x} = 5 cos^{3} (3 x + 6)

What is maclaurin e^{2x^3} ?
The answer to maclaurin e^{2x^3} is 1+2x^3+2x^6+4/3 x^9+\ldots