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}}

x \to 0 lim (\frac{x}{0})

x \to - 2 lim (x^{2} + 3 x - 7)

\frac{\partial}{\partial x} (- x y z^{2} + 1)

f (x) = - e^{x} x + 3 e^{x}

\int \frac{- 1}{4} e^{- 4 x} 3 x^{2} d x

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}