What is taylor (1+x)^{1/x} ?

The answer to taylor (1+x)^{1/x} is 2+(-2ln(2)+1)(x-1)+(2ln^2(2)+2ln(2)-2)/2 (x-1)^2+\ldots

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taylor (1+x)^{1/x}

taylor

(1 + x)^{\frac{1}{x}}

2 + (- 2 ln (2) + 1) (x - 1) + \frac{2 ln ^{2} ( 2 ) + 2 ln ( 2 ) - 2}{2} (x - 1)^{2} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

= 2 + (- 2 ln (2) + 1) (x - 1) + \frac{2 ln ^{2} ( 2 ) + 2 ln ( 2 ) - 2}{2} (x - 1)^{2} + \dots

y^{^{'''}} + 10 y^{^{''}} + 25 y^{^{'}} = 0

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

f (x) = 2 - 3 x^{2}, at x = - 2

\frac{d y}{d x} = y^{2} - 2 y

\frac{d}{d x} (e^{2 x} \cdot sin (x^{2} + 5))

What is taylor (1+x)^{1/x} ?
The answer to taylor (1+x)^{1/x} is 2+(-2ln(2)+1)(x-1)+(2ln^2(2)+2ln(2)-2)/2 (x-1)^2+\ldots