What is maclaurin ln(1+3x) ?

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

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maclaurin ln(1+3x)

maclaurin

ln (1 + 3 x)

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

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

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

\frac{d}{d x} (\frac{1}{3} x^{3} + \frac{1}{4 x})

y^{^{'}} = 2 + 3 x + 8 y + 12 x y

- 3 x^{\frac{1}{2}} + x^{\frac{3}{2}}

\frac{d}{d x} (e^{- 2 x} y)

\frac{d}{d x} (sec^{2} (2 x) - tan^{2} (2 x))

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