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

The tangent of x/(1+x^2),\at x=2 is y=-3/25 x+16/25

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

tangent of

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

y = - \frac{3}{25} x + \frac{16}{25}

Solution steps

Find the tangent point:

(2, \frac{2}{5})

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 = 2 : m = - \frac{3}{25}

Find the line with slope m=

- \frac{3}{25}

and passing through

(2, \frac{2}{5}) : y = - \frac{3}{25} x + \frac{16}{25}

y = - \frac{3}{25} x + \frac{16}{25}

\int 0.02 e^{x} d x

2 x y^{2} + 4 = 2 (3 - x^{2} y) y^{^{'}}, y (- 1) = 8

\frac{d}{d x} (x^{3} + 3 x^{2} - 144 x)

(- 1.4) (- 4.2)

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

What is the tangent of x/(1+x^2),\at x=2 ?
The tangent of x/(1+x^2),\at x=2 is y=-3/25 x+16/25