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

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

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

tangent of

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

y = - 1

Solution steps

Find the tangent point:

(0, - 1)

Find the slope of

f (x) = \frac{x ^{2} - 1}{x ^{2} + 1} : \frac{df ( x )}{d x} = \frac{4 x}{( 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

(0, - 1) : y = - 1

y = - 1

\frac{\partial}{\partial x} (- 4 cos (2 x))

\frac{d ^{2} y}{d x ^{2}} - 4 y = (x^{2} - 3) sin (2 x)

\frac{d y}{d x} + x y^{3} + \frac{y}{x} = 0

f (x) = 7 + 2 x + x^{3} (5 x^{2} + 1)^{5}

\int - x cos (nπ x) d x

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