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

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

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

tangent of

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

y = x

Solution steps

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}}

(0, 0) : m = 1

Find the line with slope m=

1

and passing through

(0, 0) : y = x

y = x

\frac{d}{d x} (\frac{3 - sec ( x )}{tan ( x )})

\frac{\partial}{\partial y} (ln (x - 2 y + 3 z))

\int_{- e^{2}}^{- e} \frac{3}{x} d x

\int \frac{1}{x ^{2} + 4 x + 3} d x

sin (x y)

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