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

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

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

tangent of

y = \frac{1}{( 1 + x ^{2} ) , at ( - 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 3 x^{15} e^{- x^{8}} d x

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lo g_{x} (2)

y = \frac{6}{x}, (16, \frac{6}{4})

y = 2 sin (x) cos (x)

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