What is the tangent of x^4-4x^3+8 ?

The tangent of x^4-4x^3+8 is y=(4a_{0}^3-12a_{0}^2)x-3a_{0}^4+8a_{0}^3+8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of x^4-4x^3+8

tangent of

x^{4} - 4 x^{3} + 8

y = (4 a_{0}^{3} - 12 a_{0}^{2}) x - 3 a_{0}^{4} + 8 a_{0}^{3} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{4} - 4 a_{0}^{3} + 8)

Find the slope of

y = x^{4} - 4 x^{3} + 8 : \frac{d y}{d x} = 4 x^{3} - 12 x^{2}

Compute the slope of

y = x^{4} - 4 x^{3} + 8

x = a_{0} : m = 4 a_{0}^{3} - 12 a_{0}^{2}

Find the line with slope m=

4 a_{0}^{3} - 12 a_{0}^{2}

and passing through

(a_{0}, a_{0}^{4} - 4 a_{0}^{3} + 8) : y = (4 a_{0}^{3} - 12 a_{0}^{2}) x - 3 a_{0}^{4} + 8 a_{0}^{3} + 8

y = (4 a_{0}^{3} - 12 a_{0}^{2}) x - 3 a_{0}^{4} + 8 a_{0}^{3} + 8

x + y^{2} = 56, x = y

\frac{\partial}{\partial y} (4 x y)

\int \frac{x ^{2} - 1}{2 x - 1} d x

\int x^{4} e^{6 x} d x

\frac{d}{d x} (\frac{x ^{3} - 1}{x})

What is the tangent of x^4-4x^3+8 ?
The tangent of x^4-4x^3+8 is y=(4a_{0}^3-12a_{0}^2)x-3a_{0}^4+8a_{0}^3+8