What is taylor ln(1+2*t),2 ?

The answer to taylor ln(1+2*t),2 is ln(5)+2/5 (t-2)-2/25 (t-2)^2+8/375 (t-2)^3-4/625 (t-2)^4+\ldots

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taylor ln(1+2*t),2

taylor

ln (1 + 2 \cdot t), 2

ln (5) + \frac{2}{5} (t - 2) - \frac{2}{25} (t - 2)^{2} + \frac{8}{375} (t - 2)^{3} - \frac{4}{625} (t - 2)^{4} + \dots

Solution steps

Apply the Taylor Formula

= ln (5) + \frac{\frac{d}{d t} ( ln ( 1 + 2 t ) ) ( 2 )}{1 !} (t - 2) + \frac{\frac{d ^{2}}{d t ^{2}} ( ln ( 1 + 2 t ) ) ( 2 )}{2 !} (t - 2)^{2} + \frac{\frac{d ^{3}}{d t ^{3}} ( ln ( 1 + 2 t ) ) ( 2 )}{3 !} (t - 2)^{3} + \dots

Evaluate Derivatives

= ln (5) + \frac{\frac{2}{5}}{1 !} (t - 2) + \frac{- \frac{4}{25}}{2 !} (t - 2)^{2} + \frac{\frac{16}{125}}{3 !} (t - 2)^{3} + \frac{- \frac{96}{625}}{4 !} (t - 2)^{4} + \dots

Refine

= ln (5) + \frac{2}{5} (t - 2) - \frac{2}{25} (t - 2)^{2} + \frac{8}{375} (t - 2)^{3} - \frac{4}{625} (t - 2)^{4} + \dots

\frac{d}{d x} (\frac{4 x - 1}{x ^{2}})

\frac{\partial}{\partial x} (- 6 y^{2} + 12 x y)

\frac{\partial}{\partial x} (\frac{( p ( q , x ) - c q )}{x})

\frac{\partial}{\partial y} (8 z e^{x yz})

f (t) = 4 t - 10

What is taylor ln(1+2*t),2 ?
The answer to taylor ln(1+2*t),2 is ln(5)+2/5 (t-2)-2/25 (t-2)^2+8/375 (t-2)^3-4/625 (t-2)^4+\ldots