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

The tangent of y=(x^2-1)/(x^2+x+1),\at x=1 is y= 2/3 x-2/3

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

tangent of

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

y = \frac{2}{3} x - \frac{2}{3}

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

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

Compute the slope of

y = \frac{x ^{2} - 1}{x ^{2} + x + 1}

x = 1 : m = \frac{2}{3}

Find the line with slope m=

\frac{2}{3}

and passing through

(1, 0) : y = \frac{2}{3} x - \frac{2}{3}

y = \frac{2}{3} x - \frac{2}{3}

\int 4 - 4 sin (x) + sin^{2} (x) d x

\int 64 - 3 x^{2} d x

\frac{\partial}{\partial x} (x^{2} e^{2})

x \to 1 lim (\frac{a x}{3})

y^{^{'}} = \frac{( 4 x ^{3} + 1 )}{2 y - 6}

What is the tangent of y=(x^2-1)/(x^2+x+1),\at x=1 ?
The tangent of y=(x^2-1)/(x^2+x+1),\at x=1 is y= 2/3 x-2/3