What is taylor ln(2x),1 ?

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

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taylor ln(2x),1

taylor

ln (2 x), 1

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= ln (2) + \frac{1}{1 !} (x - 1) + \frac{- 1}{2 !} (x - 1)^{2} + \frac{2}{3 !} (x - 1)^{3} + \frac{- 6}{4 !} (x - 1)^{4} + \dots

Refine

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

\frac{\partial}{\partial x} (e^{x} cos (θ))

f (x) = x^{4} - 6 x^{2} - x, at x = 1, y = 6

\int_{0}^{1} x^{2} (2 (1 - x)) d x

\frac{d}{d x} (2 x (6 x - 18)^{6})

\int \frac{2}{1 - e ^{x}} d x

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