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}

\frac{\partial}{\partial y} (x \cdot sin (x y))

y = ln (e^{- x} + x e^{- x})

\int 2 cos^{3} (5 x) d x

\int (4 x - 3)^{7} d x

f (x) = cos^{2} (e^{c o s^{2} (x)})

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