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

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

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

tangent of

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

y = - 1

Solution steps

Find the tangent point:

(1, - 1)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 0

Find the line with slope m=

0

and passing through

(1, - 1) : y = - 1

y = - 1

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

\int 1 + e^{x} y + x e^{x} y d x

x \to 1 lim (3 x^{2} + 1)

\int (2 - y) d y

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

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