What is the tangent of y=(x^2-1) ?

The tangent of y=(x^2-1) is y=2a_{0}x-a_{0}^2-1

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tangent of y=(x^2-1)

tangent of

y = (x^{2} - 1)

y = 2 a_{0} x - a_{0}^{2} - 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 1)

Find the slope of

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

Compute the slope of

y = (x^{2} - 1)

x = a_{0} : m = 2 a_{0}

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} - 1) : y = 2 a_{0} x - a_{0}^{2} - 1

y = 2 a_{0} x - a_{0}^{2} - 1

f (x) = 5 + 4 x^{2} - 2 x^{3}, at x = a

x \to \infty lim (- x + 2)

\int_{0}^{π} 3 sin^{2} (4 x) d x

\int e^{x} d x

f (x) = \frac{ln ( x )}{x}

What is the tangent of y=(x^2-1) ?
The tangent of y=(x^2-1) is y=2a_{0}x-a_{0}^2-1