What is maclaurin ln(1-3y) ?

The answer to maclaurin ln(1-3y) is -3y-9/2 y^2-9y^3-81/4 y^4-243/5 y^5+\ldots

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maclaurin ln(1-3y)

maclaurin

ln (1 - 3 y)

- 3 y - \frac{9}{2} y^{2} - 9 y^{3} - \frac{81}{4} y^{4} - \frac{243}{5} y^{5} + \dots

Solution steps

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d y} ( ln ( 1 - 3 y ) ) ( 0 )}{1 !} y + \frac{\frac{d ^{2}}{d y ^{2}} ( ln ( 1 - 3 y ) ) ( 0 )}{2 !} y^{2} + \frac{\frac{d ^{3}}{d y ^{3}} ( ln ( 1 - 3 y ) ) ( 0 )}{3 !} y^{3} + \dots

Evaluate Derivatives

= 0 + \frac{- 3}{1 !} y + \frac{- 9}{2 !} y^{2} + \frac{- 54}{3 !} y^{3} + \frac{- 486}{4 !} y^{4} + \frac{- 5832}{5 !} y^{5} + \dots

Refine

= - 3 y - \frac{9}{2} y^{2} - 9 y^{3} - \frac{81}{4} y^{4} - \frac{243}{5} y^{5} + \dots

\int y cos (y^{2}) d y

\frac{d}{d x} (\frac{x \cdot sin ( x )}{1 + x ^{2}})

\int_{0}^{π} 1 - cos^{2} (x) d x

\frac{d}{d x} (8 e^{π x} + 2 e^{7 x})

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

What is maclaurin ln(1-3y) ?
The answer to maclaurin ln(1-3y) is -3y-9/2 y^2-9y^3-81/4 y^4-243/5 y^5+\ldots