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}

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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}}

Compute the slope of

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

x = a_{0} : 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}}

f (x) = \frac{e ^{x}}{x + 6}, (0, \frac{1}{6})

f (x) = (x + 2 x) e^{x}

\int \frac{13}{13 + e ^{x}} d x

\frac{d y}{d x} = 6 - 3 x^{2}

\int e^{x} \cdot x^{- 1} d x

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}