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

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

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

tangent of

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

y = x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

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

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 0) : y = x

y = x

\frac{d}{d x} (- x^{2} + 6)

x \to 5 - lim (- \frac{2}{5 - x})

\frac{\partial}{\partial x} (50 y + 3250 - 100 x)

\int (1 + 2 x^{2}) e^{- 2 x} d x

(x^{3} + y^{3}) d x + (3 x y^{2}) d y = 0

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