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

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

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

tangent of

f (x) = \frac{x}{1 + x ^{2}}, 0, at x = 0

y = x

Solution steps

Find the tangent point:

(0, 0)

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}}

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 0) : y = x

y = x

\frac{d}{d x} (\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 7)

(sec (x)) \frac{d y}{d x} + y = 2

\int e^{1 x} d x

\frac{e ^{- s}}{s ^{1}}

f (x) = cos (3 x) ln (x^{4}), (1, 0)

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