What is taylor ln(2+x) ?

The answer to taylor ln(2+x) is ln(2)+1/2 x-1/8 x^2+1/24 x^3-1/64 x^4+\ldots

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taylor ln(2+x)

taylor

ln (2 + x)

ln (2) + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{24} x^{3} - \frac{1}{64} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= ln (2) + \frac{\frac{1}{2}}{1 !} x + \frac{- \frac{1}{4}}{2 !} x^{2} + \frac{\frac{1}{4}}{3 !} x^{3} + \frac{- \frac{3}{8}}{4 !} x^{4} + \dots

Refine

= ln (2) + \frac{1}{2} x - \frac{1}{8} x^{2} + \frac{1}{24} x^{3} - \frac{1}{64} x^{4} + \dots

- x + tan (x)

cos (\frac{1 - e ^{7 x}}{1 + e ^{7 x}})

\frac{\partial}{\partial x} (z x^{2})

f (x) = \frac{x}{x + 100}, at x = 50

\frac{d}{d x} (\frac{x - 5}{x + 1})

What is taylor ln(2+x) ?
The answer to taylor ln(2+x) is ln(2)+1/2 x-1/8 x^2+1/24 x^3-1/64 x^4+\ldots