What is the tangent of f(x)=x^2 ?

The tangent of f(x)=x^2 is y=2a_{0}x-a_{0}^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=x^2

tangent of

f (x) = x^{2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = x^{2} : \frac{df ( x )}{d x} = 2 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

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

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

x \to 3 lim (\frac{x - 2}{x + 5})

2 x^{2} y \frac{d y}{d x} + 1 + 2 x y^{2} = 0

\frac{d ^{3}}{d x ^{3}} (cot (2 x) + 3 x^{2})

\frac{d}{d x} (x^{4} - 8 x^{3} + 4 x^{2} + 48 x)

\frac{\partial}{\partial x} (x y + \frac{27}{x})