What is the tangent of y= x/(1+x^2),(3,0.3) ?

The tangent of y= x/(1+x^2),(3,0.3) is y=-2/25 x+0.54

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tangent of y= x/(1+x^2),(3,0.3)

tangent of

y = \frac{x}{1 + x ^{2}}, (3, 0.3)

y = - \frac{2}{25} x + 0.54

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{2}{25}

Find the line with slope m=

- \frac{2}{25}

and passing through

(3, 0.3) : y = - \frac{2}{25} x + 0.54

y = - \frac{2}{25} x + 0.54

x \to 0 + lim (\frac{2}{x} - \frac{2}{∣ x ∣})

\frac{d}{d x} (c x e^{3 x})

\frac{d}{d t} (sin (t) + t cos (t))

\frac{d}{d x} (ln ((6 x + 4) (3 x^{2} + 2)))

\int ln (11 x) d x

What is the tangent of y= x/(1+x^2),(3,0.3) ?
The tangent of y= x/(1+x^2),(3,0.3) is y=-2/25 x+0.54