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}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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{4 x}{x + 3}

x \to - 1 lim (ln (5 + 2 x))

x \to 0 lim (e^{\frac{5}{x}})

\frac{\partial}{\partial x} (x^{2} + y^{2} + 3)

\int arccos (8 x) 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}