What is the tangent of y=(8x)/((x^2+1)) ?

The tangent of y=(8x)/((x^2+1)) is y=(8(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(16a_{0}^3)/((a_{0)^2+1)^2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of y=(8x)/((x^2+1))

tangent of

y = \frac{8 x}{( x ^{2} + 1 )}

y = \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{16 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{8 a _{0}}{a _{0}^{2} + 1})

Find the slope of

y = \frac{8 x}{x ^{2} + 1} : \frac{d y}{d x} = \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

Find the line with slope m=

\frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{8 a _{0}}{a _{0}^{2} + 1}) : y = \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{16 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{16 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

What is the tangent of y=(8x)/((x^2+1)) ?
The tangent of y=(8x)/((x^2+1)) is y=(8(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(16a_{0}^3)/((a_{0)^2+1)^2}