What is the tangent of 3/(sqrt(x)),\at x= 1/16 ?

The tangent of 3/(sqrt(x)),\at x= 1/16 is y=-96x+18

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 3/(sqrt(x)),\at x= 1/16

tangent of

\frac{3}{x}, at x = \frac{1}{16}

y = - 96 x + 18

Solution steps

Find the tangent point:

(\frac{1}{16}, 12)

Find the slope of

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

Compute the slope of

y = \frac{3}{x}

x = \frac{1}{16} : m = - 96

Find the line with slope m=

- 96

and passing through

(\frac{1}{16}, 12) : y = - 96 x + 18

y = - 96 x + 18

\frac{d y}{d t} = y cos (3 t + 2)

x, 0, 0.5

\int e^{b} sin (t - b) d b

\int_{x}^{x} 1 d t

\int_{0}^{1} \frac{62}{4 y - 1} d y

What is the tangent of 3/(sqrt(x)),\at x= 1/16 ?
The tangent of 3/(sqrt(x)),\at x= 1/16 is y=-96x+18