What is the tangent of y=4x^2 ?

The tangent of y=4x^2 is y=8a_{0}x-4a_{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=4x^2

tangent of

y = 4 x^{2}

y = 8 a_{0} x - 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2})

Find the slope of

y = 4 x^{2} : \frac{d y}{d x} = 8 x

Compute the slope of

y = 4 x^{2}

x = a_{0} : m = 8 a_{0}

Find the line with slope m=

8 a_{0}

and passing through

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

y = 8 a_{0} x - 4 a_{0}^{2}

x \to - 2 lim ((x^{2} - 1) (7 x + 3))

\frac{1}{s ^{2} ( s - 1 )}

\int (3 e^{x} + 5 sec^{2} (x)) d x

\frac{\partial}{\partial y} (\frac{2 y}{x + z})

\frac{d}{d x} (8 (x^{3} - x)^{4})