What is the tangent of (-6x)/(x^2+1) ?

The tangent of (-6x)/(x^2+1) is y=-(6(-a_{0}^2+1))/((a_{0)^2+1)^2}x-(12a_{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

Popular >

tangent of (-6x)/(x^2+1)

tangent of

\frac{- 6 x}{x ^{2} + 1}

y = - \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{12 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{6 a _{0}}{a _{0}^{2} + 1})

Find the slope of

y = \frac{- 6 x}{x ^{2} + 1} : \frac{d y}{d x} = - \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

y = \frac{- 6 x}{x ^{2} + 1}

x = a_{0} : m = - \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

Find the line with slope m=

- \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

and passing through

(a_{0}, - \frac{6 a _{0}}{a _{0}^{2} + 1}) : y = - \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{12 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = - \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{12 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

\frac{d}{d y} (2 y^{3})

\int \frac{x ^{3} - 2}{x} d x

f (x) = \frac{1 - 2 t}{1 + 6 t}

f (x) = \frac{5 x}{7 x ^{2} + 1}

x \to - 2.4 lim ([x])

What is the tangent of (-6x)/(x^2+1) ?
The tangent of (-6x)/(x^2+1) is y=-(6(-a_{0}^2+1))/((a_{0)^2+1)^2}x-(12a_{0}^3)/((a_{0)^2+1)^2}