What is the tangent of-(8x)/(x^2+1) ?

The tangent of-(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-(8x)/(x^2+1)

tangent of

- \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}}

\frac{d}{d x} (2 x^{2 x} (ln (x) + 1))

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

\frac{d}{d x} (x^{\frac{2}{3}} \cdot (x - 4)^{2})

\frac{d}{d x} (\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}})

\int (x (sin (x^{2}))) (cos^{8} (x^{2})) d x

What is the tangent of-(8x)/(x^2+1) ?
The tangent of-(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}