What is maclaurin e^x,x=2 ?

The answer to maclaurin e^x,x=2 is sum from n=0 to infinity of (x^n)/(n!)

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maclaurin e^x,x=2

maclaurin

e^{x}, x = 2

Solution steps

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( e ^{x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( e ^{x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( e ^{x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{1}{1 !} x + \frac{1}{2 !} x^{2} + \frac{1}{3 !} x^{3} + \frac{1}{4 !} x^{4} + \dots

The series sum representation

= n = 0 \sum \infty \frac{x ^{n}}{n !}

t an g e n t y = \frac{4 81 + x - 3}{x}

y^{^{'}} + 8 y + 6 x = 0

\int \frac{( x + 6 )}{x} d x

\frac{d}{d x} (x cos (x) + x^{2} sin (x))

\int 3 x + 1 d x

What is maclaurin e^x,x=2 ?
The answer to maclaurin e^x,x=2 is sum from n=0 to infinity of (x^n)/(n!)