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

The tangent of (2x)/((x+1)^2) is y=(2(1-a_{0}))/((a_{0)+1)^3}x+(4a_{0}^2)/((a_{0)+1)^3}

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

tangent of

\frac{2 x}{( x + 1 ) ^{2}}

y = \frac{2 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}} x + \frac{4 a _{0}^{2}}{( a _{0} + 1 ) ^{3}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

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

and passing through

(a_{0}, \frac{2 a _{0}}{( a _{0} + 1 ) ^{2}}) : y = \frac{2 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}} x + \frac{4 a _{0}^{2}}{( a _{0} + 1 ) ^{3}}

y = \frac{2 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}} x + \frac{4 a _{0}^{2}}{( a _{0} + 1 ) ^{3}}

\int x^{- \frac{2}{3}} d x

y^{^{'}} = \frac{( x ^{2} + y )}{x ^{3}}

x \to 0 lim (x^{2} csc^{2} (2 x))

\int 2 sin (2 x) - 2 cos (2 x) d x

e^{x^{2} + 1}, 0

What is the tangent of (2x)/((x+1)^2) ?
The tangent of (2x)/((x+1)^2) is y=(2(1-a_{0}))/((a_{0)+1)^3}x+(4a_{0}^2)/((a_{0)+1)^3}