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

The tangent of f(x)= x/(1+x^2) is y=(1-a_{0}^2)/((1+a_{0)^2)^2}x+(2a_{0}^3)/((a_{0)^2+1)^2}

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

tangent of

f (x) = \frac{x}{1 + x ^{2}}

y = \frac{1 - a _{0}^{2}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{2 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{a _{0}}{1 + a _{0}^{2}})

Find the slope of

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

Compute the slope of

f (x) = \frac{x}{1 + x ^{2}}

x = a_{0} : m = \frac{1 - a _{0}^{2}}{( 1 + a _{0}^{2} ) ^{2}}

Find the line with slope m=

\frac{1 - a _{0}^{2}}{( 1 + a _{0}^{2} ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}}{1 + a _{0}^{2}}) : y = \frac{1 - a _{0}^{2}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = \frac{1 - a _{0}^{2}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{2 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = e^{x}, y = x, x = 0, x = 1

\int \frac{e ^{2 x}}{1 - e ^{4 x}} d x

x \to 2 lim (\frac{x + 1}{x ^{2} - 4})

\int \frac{1}{5 x ^{2} + x} d x

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

What is the tangent of f(x)= x/(1+x^2) ?
The tangent of f(x)= x/(1+x^2) is y=(1-a_{0}^2)/((1+a_{0)^2)^2}x+(2a_{0}^3)/((a_{0)^2+1)^2}