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

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

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

tangent of

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

y = - \frac{1}{2}

Solution steps

Find the tangent point:

(- 1, - \frac{1}{2})

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 = - 1 : m = 0

Find the line with slope m=

0

and passing through

(- 1, - \frac{1}{2}) : y = - \frac{1}{2}

y = - \frac{1}{2}

y = 2 - x, y = 4 - x^{2}

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

y^{^{''}} + 2 y + y = x e^{- x}

\int - \frac{1}{3} e^{3 x} d x

\int 2 y sin (y) d y

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