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

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

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

tangent of

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

y = x

Solution steps

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}}

(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{1}{3} x^{3} + 1)

\frac{d y}{d x} = 9 x^{2} y^{2}

\frac{d}{d x} (sin (x) + 10 tan (x))

y^{^{'}} + 0.003 y - 0.105 = 0

\frac{\partial}{\partial z} (yz ln (x y))

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