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

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

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tangent of 1/((1+x^2),(1, 1/2))

tangent of

\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}

- 3 cos (x), at x = \frac{π}{2}

y^{^{''}} - 14 y^{^{'}} + 49 y = e^{8 x} (x^{2} - 9 x - 2)

y^{^{'}} = 1 + e^{y - x + 5}

\int (x^{2} - 4)^{2} (2 x) d x

\int 13 sin^{3} (x co) s^{2} x d x

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