What is the tangent of y= x/(1+x^2),\at x=0 ?

The tangent of y= x/(1+x^2),\at x=0 is y=x

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tangent of y= x/(1+x^2),\at x=0

tangent of

y = \frac{x}{1 + x ^{2}}, at x = 0

y = x

Solution steps

Find the tangent point:

(0, 0)

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 = 1

Find the line with slope m=

1

and passing through

(0, 0) : y = x

y = x

\frac{d y}{d t} = 3 y - 1

e^{x} \cdot sin (x)

x \to 2 lim (- 16 x^{2} + 76 x)

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

\int \frac{1}{( 1 - y ) ^{2}} d y

What is the tangent of y= x/(1+x^2),\at x=0 ?
The tangent of y= x/(1+x^2),\at x=0 is y=x