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

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

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tangent of f(x)=(2x)/((x+1)^2),\at x=0

tangent of

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

y = 2 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

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

x = 0 : m = 2

Find the line with slope m=

2

and passing through

(0, 0) : y = 2 x

y = 2 x

(2.4) (6.6)

\frac{d}{d x} (2 x^{3} y)

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

\int \frac{1}{x - 3} d x

y = 12 - x, y = x, y = 1

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