What is taylor 4x^4+3x^3+2x^2+x+1,1 ?

The answer to taylor 4x^4+3x^3+2x^2+x+1,1 is 11+30(x-1)+35(x-1)^2+19(x-1)^3+4(x-1)^4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

taylor 4x^4+3x^3+2x^2+x+1,1

taylor

4 x^{4} + 3 x^{3} + 2 x^{2} + x + 1, 1

11 + 30 (x - 1) + 35 (x - 1)^{2} + 19 (x - 1)^{3} + 4 (x - 1)^{4}

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 11 + \frac{30}{1 !} (x - 1) + \frac{70}{2 !} (x - 1)^{2} + \frac{114}{3 !} (x - 1)^{3} + \frac{96}{4 !} (x - 1)^{4}

Refine

= 11 + 30 (x - 1) + 35 (x - 1)^{2} + 19 (x - 1)^{3} + 4 (x - 1)^{4}

\frac{\partial}{\partial x} (- 5 y e^{x})

\frac{( s + 2 )}{s ( s ^{2} + 9 )}

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

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

\frac{d}{d x} (\frac{π}{4})

What is taylor 4x^4+3x^3+2x^2+x+1,1 ?
The answer to taylor 4x^4+3x^3+2x^2+x+1,1 is 11+30(x-1)+35(x-1)^2+19(x-1)^3+4(x-1)^4