What is the tangent of y=(5x)/((x+1)^2),(0,0) ?

The tangent of y=(5x)/((x+1)^2),(0,0) is y=5x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=(5x)/((x+1)^2),(0,0)

tangent of

y = \frac{5 x}{( x + 1 ) ^{2}}, (0, 0)

y = 5 x

Solution steps

Find the slope of

y = \frac{5 x}{( x + 1 ) ^{2}} : \frac{d y}{d x} = \frac{5 ( 1 - x )}{( x + 1 ) ^{3}}

Compute the slope of

y = \frac{5 x}{( x + 1 ) ^{2}}

(0, 0) : m = 5

Find the line with slope m=

5

and passing through

(0, 0) : y = 5 x

y = 5 x

x y^{^{'}} + y^{2} = 0

\int - x d x

\frac{\partial}{\partial x} (y \cdot arctan (\frac{x}{z}))

\int (\frac{x ^{3} - 1}{x - 1}) d x

\frac{d}{d x} (sec (\frac{π}{6}))

What is the tangent of y=(5x)/((x+1)^2),(0,0) ?
The tangent of y=(5x)/((x+1)^2),(0,0) is y=5x