What is the tangent of sqrt(x+2),\at x=4 ?

The tangent of sqrt(x+2),\at x=4 is y= 1/(2sqrt(6))x+sqrt(6)-sqrt(2/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of sqrt(x+2),\at x=4

tangent of

x + 2, at x = 4

y = \frac{1}{2 6} x + 6 - \frac{2}{3}

Solution steps

Find the tangent point:

(4, 6)

Find the slope of

y = x + 2 : \frac{d y}{d x} = \frac{1}{2 x + 2}

Compute the slope of

y = x + 2

x = 4 : m = \frac{1}{2 6}

Find the line with slope m=

\frac{1}{2 6}

and passing through

(4, 6) : y = \frac{1}{2 6} x + 6 - \frac{2}{3}

y = \frac{1}{2 6} x + 6 - \frac{2}{3}

\int (6 x^{2} - 5 x + 4) d x

\frac{d}{d t} (x e^{\frac{y}{z ( t )}})

y = ln (e^{x})

n = 1000 \sum \infty \frac{1}{n}

12 - x, x, 1

What is the tangent of sqrt(x+2),\at x=4 ?
The tangent of sqrt(x+2),\at x=4 is y= 1/(2sqrt(6))x+sqrt(6)-sqrt(2/3)