What is taylor ln(4x),1 ?

The answer to taylor ln(4x),1 is 2ln(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(4x),1

taylor

ln (4 x), 1

2 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

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

Evaluate Derivatives

= 2 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

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

\int \frac{1}{( x ^{2} - 64 ) ^{2}} d x

\int \frac{2 - x}{2 + x} d x

\frac{\partial}{\partial x} (x y - x sin (y))

\frac{( t ^{2} + 1 ) ^{\frac{3}{2}}}{3}^{^{'}}

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

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