What is the tangent of 1/(x^2),\at x=3 ?

The tangent of 1/(x^2),\at x=3 is y=-2/27 x+1/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 1/(x^2),\at x=3

tangent of

\frac{1}{x ^{2}}, at x = 3

y = - \frac{2}{27} x + \frac{1}{3}

Solution steps

Find the tangent point:

(3, \frac{1}{9})

Find the slope of

y = \frac{1}{x ^{2}} : \frac{d y}{d x} = - \frac{2}{x ^{3}}

Compute the slope of

y = \frac{1}{x ^{2}}

x = 3 : m = - \frac{2}{27}

Find the line with slope m=

- \frac{2}{27}

and passing through

(3, \frac{1}{9}) : y = - \frac{2}{27} x + \frac{1}{3}

y = - \frac{2}{27} x + \frac{1}{3}

\frac{d y}{d x} 3 x^{3} + 3 y^{3} - 5 x y = 0

4 x^{6} e^{x}

\int \frac{t}{a t ^{2} + b} d t

\frac{d ^{2}}{d x ^{2}} (3 e^{x} cos (x))

(- 3, 2), (6, 3)

What is the tangent of 1/(x^2),\at x=3 ?
The tangent of 1/(x^2),\at x=3 is y=-2/27 x+1/3