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}}

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 = 0

Find the line with slope m=

0

and passing through

(1, 1) : y = 1

y = 1

y^{^{''}} = - 3 y

\frac{\partial}{\partial x} (sec (x y) + tan (x y))

x \to 4 + lim (\frac{2}{x - 4})

\int 3 x^{- 5} d x

\int (e^{x} + 2) d 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