What is the tangent of y=(2x)/((x+1)^2),(0,0) ?

The tangent of y=(2x)/((x+1)^2),(0,0) is y=2x

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tangent of y=(2x)/((x+1)^2),(0,0)

tangent of

y = \frac{2 x}{( x + 1 ) ^{2}}, (0, 0)

y = 2 x

Solution steps

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}}

(0, 0) : m = 2

Find the line with slope m=

2

and passing through

(0, 0) : y = 2 x

y = 2 x

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

\frac{ln ( x )}{5 x ^{2}}

\frac{\partial}{\partial x} (e^{2 x + 7 y})

y^{^{''}} + 6 y^{^{'}} + 9 y = 6 t^{2} e^{- 3 t}

\frac{\partial}{\partial x} (x e^{x^{2} - y})

What is the tangent of y=(2x)/((x+1)^2),(0,0) ?
The tangent of y=(2x)/((x+1)^2),(0,0) is y=2x