What is the tangent of y=sqrt(x-1),(10,3) ?

The tangent of y=sqrt(x-1),(10,3) is y= 1/6 x+4/3

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),(10,3)

tangent of

y = x - 1, (10, 3)

y = \frac{1}{6} x + \frac{4}{3}

Solution steps

Find the slope of

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

Compute the slope of

y = x - 1

(10, 3) : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

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

y = \frac{1}{6} x + \frac{4}{3}

f (x) = 2 e^{0}

\int \frac{1}{cos ^{2} ( x ) 1 + tan ( x )} d x

\frac{\partial}{\partial x} (\frac{9 x}{y})

y = x x, (9, 27)

\frac{d}{d n} (5^{n})

What is the tangent of y=sqrt(x-1),(10,3) ?
The tangent of y=sqrt(x-1),(10,3) is y= 1/6 x+4/3