What is the tangent of y= 1/((1+x^2),(-1, 1/2)) ?

The tangent of y= 1/((1+x^2),(-1, 1/2)) is

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tangent of y= 1/((1+x^2),(-1, 1/2))

tangent of

y = \frac{1}{( 1 + x ^{2} ) , ( - 1 , \frac{1}{2} )}

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

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} ) , ( - 1 , \frac{1}{2} )})

Find the slope of

y = \frac{1}{( 1 + x ^{2} ) , ( - 1 , \frac{1}{2} )} : \frac{d y}{d x} = \frac{1}{, ( - 1 , \frac{1}{2} )} (- \frac{1}{( 1 + x ^{2} ) ^{2}} \cdot 2 x)

Compute the slope of

y = \frac{1}{( 1 + x ^{2} ) , ( - 1 , \frac{1}{2} )}

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

Find the line with slope m=

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

and passing through

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

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

\int (x e^{2 x}) d x

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 5 y = 0

\int \frac{x ^{2} + 18 x - 9}{x ^{3} - 9 x} d x

y = x (x + 1)^{2}, y = 0

\int_{0}^{\frac{π}{12}} cos (4 x) d x

What is the tangent of y= 1/((1+x^2),(-1, 1/2)) ?
The tangent of y= 1/((1+x^2),(-1, 1/2)) is