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

The tangent of sqrt(x+16),\at x=1 is y= 1/(2sqrt(17))x+sqrt(17)-1/(2sqrt(17))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

x + 16, at x = 1

y = \frac{1}{2 17} x + 17 - \frac{1}{2 17}

Solution steps

Find the tangent point:

(1, 17)

Find the slope of

y = x + 16 : \frac{d y}{d x} = \frac{1}{2 x + 16}

Compute the slope of

y = x + 16

x = 1 : m = \frac{1}{2 17}

Find the line with slope m=

\frac{1}{2 17}

and passing through

(1, 17) : y = \frac{1}{2 17} x + 17 - \frac{1}{2 17}

y = \frac{1}{2 17} x + 17 - \frac{1}{2 17}

x \to 7 lim (ln (x - 7))

y = e^{3 x} cos (6 x)

\frac{1}{1 + x ^{2}}, 1

\frac{4}{5 x + 1}

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 1})

What is the tangent of sqrt(x+16),\at x=1 ?
The tangent of sqrt(x+16),\at x=1 is y= 1/(2sqrt(17))x+sqrt(17)-1/(2sqrt(17))