What is the tangent of y=(5x)/(1+x^2) ?

The tangent of y=(5x)/(1+x^2) is y=(5(1-a_{0}^2))/((1+a_{0)^2)^2}x+(10a_{0}^3)/((a_{0)^2+1)^2}

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tangent of y=(5x)/(1+x^2)

tangent of

y = \frac{5 x}{1 + x ^{2}}

y = \frac{5 ( 1 - a _{0}^{2} )}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{10 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{5 a _{0}}{1 + a _{0}^{2}})

Find the slope of

y = \frac{5 x}{1 + x ^{2}} : \frac{d y}{d x} = \frac{5 ( 1 - x ^{2} )}{( 1 + 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{5 ( 1 - a _{0}^{2} )}{( 1 + a _{0}^{2} ) ^{2}}

Find the line with slope m=

\frac{5 ( 1 - a _{0}^{2} )}{( 1 + a _{0}^{2} ) ^{2}}

and passing through

(a_{0}, \frac{5 a _{0}}{1 + a _{0}^{2}}) : y = \frac{5 ( 1 - a _{0}^{2} )}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{10 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = \frac{5 ( 1 - a _{0}^{2} )}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{10 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

x \to \infty lim (12)

y = 4 x \cdot tan (x), at x = π

\int 4 sin (x) d x

\frac{d y}{d t} = 7 t e^{2 y}

\frac{\partial}{\partial x} (0.5 x arctan (\frac{y}{x}))

What is the tangent of y=(5x)/(1+x^2) ?
The tangent of y=(5x)/(1+x^2) is y=(5(1-a_{0}^2))/((1+a_{0)^2)^2}x+(10a_{0}^3)/((a_{0)^2+1)^2}