What is the tangent of y=8x^2-x^3(2.24) ?

The tangent of y=8x^2-x^3(2.24) is y=(16a_{0}-6.72a_{0}^2)x+4.48a_{0}^3-8a_{0}^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=8x^2-x^3(2.24)

tangent of

y = 8 x^{2} - x^{3} (2.24)

y = (16 a_{0} - 6.72 a_{0}^{2}) x + 4.48 a_{0}^{3} - 8 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 a_{0}^{2} - a_{0}^{3} \cdot 2.24)

Find the slope of

y = 8 x^{2} - x^{3} 2.24 : \frac{d y}{d x} = 16 x - 6.72 x^{2}

Compute the slope of

y = 8 x^{2} - x^{3} 2.24

x = a_{0} : m = 16 a_{0} - 6.72 a_{0}^{2}

Find the line with slope m=

16 a_{0} - 6.72 a_{0}^{2}

and passing through

(a_{0}, 8 a_{0}^{2} - a_{0}^{3} 2.24) : y = (16 a_{0} - 6.72 a_{0}^{2}) x + 4.48 a_{0}^{3} - 8 a_{0}^{2}

y = (16 a_{0} - 6.72 a_{0}^{2}) x + 4.48 a_{0}^{3} - 8 a_{0}^{2}

n = 0 \sum \infty \frac{( 8 ^{n} )}{n !}

y^{^{'}} + y = y^{2}

\frac{d}{d x} (30 x^{4} y^{7} - 24 x^{12} y^{3})

\int 9 x e^{15 x} d x

\frac{d}{d x} (- e^{x^{2}})

What is the tangent of y=8x^2-x^3(2.24) ?
The tangent of y=8x^2-x^3(2.24) is y=(16a_{0}-6.72a_{0}^2)x+4.48a_{0}^3-8a_{0}^2