What is the tangent of x^2-4,\at x=3 ?

The tangent of x^2-4,\at x=3 is y=6x-13

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of x^2-4,\at x=3

tangent of

x^{2} - 4, at x = 3

y = 6 x - 13

Solution steps

Find the tangent point:

(3, 5)

Find the slope of

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

Compute the slope of

y = x^{2} - 4

x = 3 : m = 6

Find the line with slope m=

6

and passing through

(3, 5) : y = 6 x - 13

y = 6 x - 13

\frac{ln ( x )}{10 x}

\frac{d}{d x} (\frac{2 x + cos ( 2 x )}{x ^{2} + sin ( 2 x )})

\frac{\partial}{\partial x} ((x^{2} + y^{2}) e^{- x})

x \to 1 lim (\frac{4}{x ^{2} - 1})

\frac{d}{d x} ((3 x + 1)^{4})