What is the tangent of f(x)=(3x)/((x+1)^2),\at x=0 ?

The tangent of f(x)=(3x)/((x+1)^2),\at x=0 is y=3x

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tangent of f(x)=(3x)/((x+1)^2),\at x=0

tangent of

f (x) = \frac{3 x}{( x + 1 ) ^{2}}, at x = 0

y = 3 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

f (x) = \frac{3 x}{( x + 1 ) ^{2}}

x = 0 : m = 3

Find the line with slope m=

3

and passing through

(0, 0) : y = 3 x

y = 3 x

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

\frac{d}{d x} (4 sin^{3} (x))

\frac{d}{d x} ((lo g_{10} (x))^{l o g_{10} (x)})

y^{^{'}} = 2 - y

3 x, x^{2} - 4

What is the tangent of f(x)=(3x)/((x+1)^2),\at x=0 ?
The tangent of f(x)=(3x)/((x+1)^2),\at x=0 is y=3x