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

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

tangent of

f (x) = \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

f (x) = \frac{8 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \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}}

(4 - t^{2}) y^{^{'}} + 2 t y = 3 t^{2}

\int \frac{1}{2} t^{2} d t

\frac{\partial}{\partial x} (x e^{y} + y e^{- x})

\int (5 x + 7 y) d x

f (x) = 2 sin (x) + x, - π, π

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