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

The tangent of y=(x^2)/(2-x^2) is y=(4a_{0})/((2-a_{0)^2)^2}x-(a_{0}^2(a_{0}^2+2))/((-a_{0)^2+2)^2}

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tangent of y=(x^2)/(2-x^2)

tangent of

y = \frac{x ^{2}}{2 - x ^{2}}

y = \frac{4 a _{0}}{( 2 - a _{0}^{2} ) ^{2}} x - \frac{a _{0}^{2} ( a _{0}^{2} + 2 )}{( - a _{0}^{2} + 2 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = \frac{x ^{2}}{2 - x ^{2}} : \frac{d y}{d x} = \frac{4 x}{( 2 - x ^{2} ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{4 a _{0}}{( 2 - a _{0}^{2} ) ^{2}}

Find the line with slope m=

\frac{4 a _{0}}{( 2 - a _{0}^{2} ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}^{2}}{2 - a _{0}^{2}}) : y = \frac{4 a _{0}}{( 2 - a _{0}^{2} ) ^{2}} x - \frac{a _{0}^{2} ( a _{0}^{2} + 2 )}{( - a _{0}^{2} + 2 ) ^{2}}

y = \frac{4 a _{0}}{( 2 - a _{0}^{2} ) ^{2}} x - \frac{a _{0}^{2} ( a _{0}^{2} + 2 )}{( - a _{0}^{2} + 2 ) ^{2}}

y^{^{''}} + 4 y = x sin (2 x)

\frac{d}{d x} (4 2 x - 1)

x \to 3 lim (\frac{3 x}{x - 3})

\int \frac{x + 1}{x ^{2} + 3} d x

\frac{d}{d x} (\frac{3 x ^{3} + 9 x ^{2}}{x})

What is the tangent of y=(x^2)/(2-x^2) ?
The tangent of y=(x^2)/(2-x^2) is y=(4a_{0})/((2-a_{0)^2)^2}x-(a_{0}^2(a_{0}^2+2))/((-a_{0)^2+2)^2}