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

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

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

normal of

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

x = - 1

Solution steps

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = \frac{x}{x ^{2} + 1}

x = - 1 : m = 0

Compute the slope of the perpendicular line:

m_{p} = \infty

Find the line with slope m=

\infty

and passing through

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

x = - 1

\frac{\partial}{\partial x} (2 x \cdot y - \frac{1}{2} x^{2})

f (t) = 6 sin (t)

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

\frac{d}{d x} (x^{2} + 7 x + 12)

\int \frac{1}{cos ( x ) sin ( x ) + cos ^{2} ( x )} d x

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