What is the tangent of y=sqrt(1-4x),\at x=-1 ?

The tangent of y=sqrt(1-4x),\at x=-1 is y=-2/(sqrt(5))x+sqrt(5)-2/(sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=sqrt(1-4x),\at x=-1

tangent of

y = 1 - 4 x, at x = - 1

y = - \frac{2}{5} x + 5 - \frac{2}{5}

Solution steps

Find the tangent point:

(- 1, 5)

Find the slope of

y = 1 - 4 x : \frac{d y}{d x} = - \frac{2}{1 - 4 x}

Compute the slope of

y = 1 - 4 x

x = - 1 : m = - \frac{2}{5}

Find the line with slope m=

- \frac{2}{5}

and passing through

(- 1, 5) : y = - \frac{2}{5} x + 5 - \frac{2}{5}

y = - \frac{2}{5} x + 5 - \frac{2}{5}

\int \frac{1}{( 16 x ^{2} + 9 ) ^{2}} d x

\frac{7}{x}, at x = 6

\frac{\partial}{\partial x} (sin (- x) cos (8 y))

2 tan (cos (x))

What is the tangent of y=sqrt(1-4x),\at x=-1 ?
The tangent of y=sqrt(1-4x),\at x=-1 is y=-2/(sqrt(5))x+sqrt(5)-2/(sqrt(5))