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

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

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

tangent of

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

y = x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

y = \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

\int 36 x e^{- 9 x} d x

\frac{2 x + 1}{x + 5} \cdot (3 x - 1)

e^{16 x}

\frac{d}{d x} (ln (x^{2} + 3))

\int (1) d x

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