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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

x, at x = 1

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

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

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

Compute the slope of

y = x

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

Find the line with slope m=

\frac{1}{2}

and passing through

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

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

\int x (x - 3) d x

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

f (x) = - \frac{7}{( x + 4 ) ^{2}}

f (t) = (1) (t^{3})

\frac{d}{d x} (\frac{1}{16 x ^{4}})

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