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

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

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

tangent of

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

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

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

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

and passing through

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

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

\frac{d}{d x} (4 - \frac{20}{x ^{3}} + \frac{3}{x ^{5}})

f (x) = x^{2} + x, at (x = 1)

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

3 x y^{2} y^{^{'}} = 3 x^{4} + y^{3}

(3.7) (6.12)

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