What is taylor (x^2+1)e^{x-1},1 ?

The answer to taylor (x^2+1)e^{x-1},1 is 2+4(x-1)+4(x-1)^2+7/3 (x-1)^3+11/12 (x-1)^4+\ldots

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taylor (x^2+1)e^{x-1},1

taylor

(x^{2} + 1) e^{x - 1}, 1

2 + 4 (x - 1) + 4 (x - 1)^{2} + \frac{7}{3} (x - 1)^{3} + \frac{11}{12} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 2 + \frac{4}{1 !} (x - 1) + \frac{8}{2 !} (x - 1)^{2} + \frac{14}{3 !} (x - 1)^{3} + \frac{22}{4 !} (x - 1)^{4} + \dots

Refine

= 2 + 4 (x - 1) + 4 (x - 1)^{2} + \frac{7}{3} (x - 1)^{3} + \frac{11}{12} (x - 1)^{4} + \dots

\frac{d y}{d x} = \frac{( 4 sec ( y ) )}{( x + 3 ) ^{2}}

\frac{d}{d x} (\frac{( x ^{4} - 5 x ^{3} + x )}{x ^{2}})

\frac{d}{d x} (e^{4 x} tan (x^{5} - 4 x^{4}))

\int 9 sin^{2} (x co) s^{2} x d x

y = x^{7} e^{x}

What is taylor (x^2+1)e^{x-1},1 ?
The answer to taylor (x^2+1)e^{x-1},1 is 2+4(x-1)+4(x-1)^2+7/3 (x-1)^3+11/12 (x-1)^4+\ldots