What is the tangent of y=-(4x)/(x^2+1) ?

The tangent of y=-(4x)/(x^2+1) is y=-(4(-a_{0}^2+1))/((a_{0)^2+1)^2}x-(8a_{0}^3)/((a_{0)^2+1)^2}

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tangent of y=-(4x)/(x^2+1)

tangent of

y = - \frac{4 x}{x ^{2} + 1}

y = - \frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{8 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{4 a _{0}}{a _{0}^{2} + 1})

Find the slope of

y = - \frac{4 x}{x ^{2} + 1} : \frac{d y}{d x} = - \frac{4 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

y = - \frac{4 x}{x ^{2} + 1}

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

Find the line with slope m=

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

and passing through

(a_{0}, - \frac{4 a _{0}}{a _{0}^{2} + 1}) : y = - \frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{8 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = - \frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{8 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

\frac{arccos ( x )}{x + 1}

\int x^{2} + 5 x + 2 d x

\frac{d y}{d t} = e^{- t} - y

\frac{d}{d x} (7 e^{x} + x)

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

What is the tangent of y=-(4x)/(x^2+1) ?
The tangent of y=-(4x)/(x^2+1) is y=-(4(-a_{0}^2+1))/((a_{0)^2+1)^2}x-(8a_{0}^3)/((a_{0)^2+1)^2}