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}}

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{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}}

f (x) = x^{2.4} + e^{2.4}

\frac{\partial}{\partial x} (5 x^{4} y - 3 y sin (x))

(4.4, 8997.96), (107.3, 21421.44)

f (x) = e^{l n (s i n (x^{2}))} 2 x + 1

y^{^{''}} - (\frac{3}{x}) y^{^{'}} - (\frac{5}{x ^{2}}) y = 0

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}