What is the tangent of y= x/((1+x^2),\at (3,0.3)) ?

The tangent of y= x/((1+x^2),\at (3,0.3)) is

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tangent of y= x/((1+x^2),\at (3,0.3))

tangent of

y = \frac{x}{( 1 + x ^{2} ) , at ( 3 , 0.3 )}

y = \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{a _{0}}{( 1 + a _{0}^{2} ) , ( 3 , 0.3 )} - \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} a_{0}

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} ) , ( 3 , 0.3 )})

Find the slope of

y = \frac{x}{( 1 + x ^{2} ) , ( 3 , 0.3 )} : \frac{d y}{d x} = \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + x ^{2} ) - 2 xx}{( 1 + x ^{2} ) ^{2}}

Compute the slope of

y = \frac{x}{( 1 + x ^{2} ) , ( 3 , 0.3 )}

x = a_{0} : m = \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

Find the line with slope m=

\frac{1}{, ( 3 , 0.3 )} \frac{1 ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}}{( 1 + a _{0}^{2} ) , ( 3 , 0.3 )}) : y = \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{a _{0}}{( 1 + a _{0}^{2} ) , ( 3 , 0.3 )} - \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} a_{0}

y = \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{a _{0}}{( 1 + a _{0}^{2} ) , ( 3 , 0.3 )} - \frac{1}{, ( 3 , 0.3 )} \cdot \frac{1 \cdot ( 1 + a _{0}^{2} ) - 2 a _{0} a _{0}}{( 1 + a _{0}^{2} ) ^{2}} a_{0}

\frac{d}{d x} (1 + 4 sin (x))

\int - 2 y - y^{2} + 2 d y

f (x) = 1 - 7 x^{2}, (- 2, - 27)

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

y^{^{''}} + 13 y^{^{'}} + 42 y = 0

What is the tangent of y= x/((1+x^2),\at (3,0.3)) ?
The tangent of y= x/((1+x^2),\at (3,0.3)) is