What is the tangent of y=sqrt(7x+1),(5,6) ?

The tangent of y=sqrt(7x+1),(5,6) is y= 7/12 x+37/12

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=sqrt(7x+1),(5,6)

tangent of

y = 7 x + 1, (5, 6)

y = \frac{7}{12} x + \frac{37}{12}

Solution steps

Find the slope of

y = 7 x + 1 : \frac{d y}{d x} = \frac{7}{2 7 x + 1}

Compute the slope of

y = 7 x + 1

(5, 6) : m = \frac{7}{12}

Find the line with slope m=

\frac{7}{12}

and passing through

(5, 6) : y = \frac{7}{12} x + \frac{37}{12}

y = \frac{7}{12} x + \frac{37}{12}

x \to 0 lim ((1 + 3 x)^{\frac{4}{x}})

\frac{\partial}{\partial x} ((y - x^{2}) (2 y^{2} + x))

\int \frac{5}{36 + x ^{2}} d x

\frac{\partial}{\partial y} (x y (1 - x - y))

y = x^{4} + 2 e^{x}

What is the tangent of y=sqrt(7x+1),(5,6) ?
The tangent of y=sqrt(7x+1),(5,6) is y= 7/12 x+37/12