What is the tangent of y= 2/(1+x^2),(3,0.2) ?

The tangent of y= 2/(1+x^2),(3,0.2) is y=-3/25 x+0.56

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tangent of y= 2/(1+x^2),(3,0.2)

tangent of

y = \frac{2}{1 + x ^{2}}, (3, 0.2)

y = - \frac{3}{25} x + 0.56

Solution steps

Find the slope of

y = \frac{2}{1 + x ^{2}} : \frac{d y}{d x} = - \frac{4 x}{( 1 + x ^{2} ) ^{2}}

Compute the slope of

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

(3, 0.2) : m = - \frac{3}{25}

Find the line with slope m=

- \frac{3}{25}

and passing through

(3, 0.2) : y = - \frac{3}{25} x + 0.56

y = - \frac{3}{25} x + 0.56

x (\frac{d y}{d x}) - y = x^{5} + 1

f (x) = x^{2} - 5 x

x \to 0 + lim (8 sin (x) ln (x))

\int \frac{2 x + 1}{( x ^{2} + x + 1 ) ^{\frac{3}{2}}} d x

\int \frac{9 x ^{2} - 26 x - 5}{3 x ^{2} - 5 x - 2} d x

What is the tangent of y= 2/(1+x^2),(3,0.2) ?
The tangent of y= 2/(1+x^2),(3,0.2) is y=-3/25 x+0.56