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

The tangent of y=(-2x)/(x^2+1) is y=-(2(-a_{0}^2+1))/((a_{0)^2+1)^2}x-(4a_{0}^3)/((a_{0)^2+1)^2}

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

tangent of

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

y = - \frac{2 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{4 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{2 a _{0}}{a _{0}^{2} + 1})

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

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

and passing through

(a_{0}, - \frac{2 a _{0}}{a _{0}^{2} + 1}) : y = - \frac{2 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{4 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = - \frac{2 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{4 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

\frac{d y}{d x} + 6 x^{5} y = x^{5}

\frac{\partial}{\partial z} (10 e^{x} sin (y) z^{3})

\frac{d}{d x} ((A x^{2} + B x + C) (e^{4 x}))

y^{^{'}} = \frac{2 y ^{4} - 2}{4 x y ^{3}}

n = 3 \sum \infty 1 + \frac{1}{n ^{2}}

What is the tangent of y=(-2x)/(x^2+1) ?
The tangent of y=(-2x)/(x^2+1) is y=-(2(-a_{0}^2+1))/((a_{0)^2+1)^2}x-(4a_{0}^3)/((a_{0)^2+1)^2}