What is the tangent of y= 8/(sqrt(x)),(16, 8/4) ?

The tangent of y= 8/(sqrt(x)),(16, 8/4) is y=-1/16 x+3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y= 8/(sqrt(x)),(16, 8/4)

tangent of

y = \frac{8}{x}, (16, \frac{8}{4})

y = - \frac{1}{16} x + 3

Solution steps

Find the slope of

y = \frac{8}{x} : \frac{d y}{d x} = - \frac{4}{x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{8}{x}

(16, \frac{8}{4}) : m = - \frac{1}{16}

Find the line with slope m=

- \frac{1}{16}

and passing through

(16, \frac{8}{4}) : y = - \frac{1}{16} x + 3

y = - \frac{1}{16} x + 3

\frac{d}{d x} (\frac{2}{( 1 - x ) ^{3}})

\int_{0}^{1} 5 (1 + x)^{6} d x

\frac{\partial}{\partial x} (x y e^{x z})

\frac{d}{d x} (arcsec (\frac{1}{x}))

x \to - 4 lim (3 x^{2} - 4 x + 8)

What is the tangent of y= 8/(sqrt(x)),(16, 8/4) ?
The tangent of y= 8/(sqrt(x)),(16, 8/4) is y=-1/16 x+3