What is the tangent of y= 6/(sqrt(x)),(4, 6/2) ?

The tangent of y= 6/(sqrt(x)),(4, 6/2) is y=-3/8 x+9/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y= 6/(sqrt(x)),(4, 6/2)

tangent of

y = \frac{6}{x}, (4, \frac{6}{2})

y = - \frac{3}{8} x + \frac{9}{2}

Solution steps

Find the slope of

y = \frac{6}{x} : \frac{d y}{d x} = - \frac{3}{x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{6}{x}

(4, \frac{6}{2}) : m = - \frac{3}{8}

Find the line with slope m=

- \frac{3}{8}

and passing through

(4, \frac{6}{2}) : y = - \frac{3}{8} x + \frac{9}{2}

y = - \frac{3}{8} x + \frac{9}{2}

\int \frac{e ^{x}}{3} d x

\int \frac{6 - e ^{x}}{e ^{5 x}} d x

e^{x}, e^{- 4 x}, 0, ln (4)

f (x) = 3 x^{5}

\frac{d}{d x} (e^{- x} - 1)

What is the tangent of y= 6/(sqrt(x)),(4, 6/2) ?
The tangent of y= 6/(sqrt(x)),(4, 6/2) is y=-3/8 x+9/2