What is taylor \sqrt[3]{1+sin(x)} ?

The answer to taylor \sqrt[3]{1+sin(x)} is 1+1/3 x-1/9 x^2+1/162 x^3-1/243 x^4+\ldots

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

taylor \sqrt[3]{1+sin(x)}

taylor

3 1 + sin (x)

1 + \frac{1}{3} x - \frac{1}{9} x^{2} + \frac{1}{162} x^{3} - \frac{1}{243} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{\frac{1}{3}}{1 !} x + \frac{- \frac{2}{9}}{2 !} x^{2} + \frac{\frac{1}{27}}{3 !} x^{3} + \frac{- \frac{8}{81}}{4 !} x^{4} + \dots

Refine

= 1 + \frac{1}{3} x - \frac{1}{9} x^{2} + \frac{1}{162} x^{3} - \frac{1}{243} x^{4} + \dots

(1 + \frac{1}{x})^{- 2}

f (x) = 3 x + 1, [0, 3]

f (x) = sin (x), at x = 0

\frac{d}{d x} (\frac{- 1}{x + c})

x y^{^{'}} + y = y^{2} ln (x)

What is taylor \sqrt[3]{1+sin(x)} ?
The answer to taylor \sqrt[3]{1+sin(x)} is 1+1/3 x-1/9 x^2+1/162 x^3-1/243 x^4+\ldots