What is taylor e^{2x},1 ?

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

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taylor e^{2x},1

taylor

e^{2 x}, 1

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= e^{2} + \frac{2 e ^{2}}{1 !} (x - 1) + \frac{4 e ^{2}}{2 !} (x - 1)^{2} + \frac{8 e ^{2}}{3 !} (x - 1)^{3} + \frac{16 e ^{2}}{4 !} (x - 1)^{4} + \dots

Refine

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

\frac{\partial}{\partial x} (x^{2} - 4 y)

y \frac{d y}{d x} - e^{y^{2}} \cdot x \cdot cos (x) = 0

\frac{d}{d x} (- \frac{50}{x ^{3}} + \frac{5}{x ^{2}})

\int 4 x^{2} sin (2 x) d x

\frac{\partial}{\partial x} (ln (1 + x^{2} + y^{2} - z^{2}))

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