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{d}{d x} (x^{\frac{1}{3}} (x + 3)^{\frac{2}{3}})

\int 7 tan (x) d x

\int \frac{1}{sin ^{5} ( x ) cos ( x )} d x

\frac{d}{d x} (- \frac{x ^{3}}{3})

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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}