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

The tangent of y=sqrt(x),(1,1) 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 y=sqrt(x),(1,1)

tangent of

y = x, (1, 1)

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

Solution steps

Find the slope of

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

Compute the slope of

y = x

(1, 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}

\frac{d}{d x} (\frac{sin ( x )}{x + 1})

x \to 0 lim (\frac{( 2 x ^{2} - 3 x )}{x})

3 \cdot (5 e^{- 5.06 s}) + 3 \cdot (2.26 \cdot 5 e^{- 5.06 s})

\int \frac{( 9 4 - x ^{2} )}{2} d x

\frac{\partial}{\partial x} (sin (2 x + y))

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