What is the tangent of f(x)= x/(1+x^2),(2,0.4) ?

The tangent of f(x)= x/(1+x^2),(2,0.4) is y=-3/25 x+0.64

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tangent of f(x)= x/(1+x^2),(2,0.4)

tangent of

f (x) = \frac{x}{1 + x ^{2}}, (2, 0.4)

y = - \frac{3}{25} x + 0.64

Solution steps

Find the slope of

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

Compute the slope of

f (x) = \frac{x}{1 + x ^{2}}

(2, 0.4) : m = - \frac{3}{25}

Find the line with slope m=

- \frac{3}{25}

and passing through

(2, 0.4) : y = - \frac{3}{25} x + 0.64

y = - \frac{3}{25} x + 0.64

\int_{0}^{2} (x + 2) - x^{2} d x

\frac{d}{d x} ((1 - sin (x))^{- \frac{1}{2}})

\frac{d}{d x} (ln (\frac{x}{3}) (x - 3))

\int \frac{361}{x ^{3} - 19 x ^{2}} d x

\frac{d}{d x} (- 6 sin (x))

What is the tangent of f(x)= x/(1+x^2),(2,0.4) ?
The tangent of f(x)= x/(1+x^2),(2,0.4) is y=-3/25 x+0.64