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

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

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tangent of f(x)=x^2-1

tangent of

f (x) = 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

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

Compute the slope of

f (x) = 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

\int \frac{x ^{3}}{x ^{4} - 2} d x

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

\frac{\partial}{\partial x} (\frac{- 4 x ^{5}}{4 y ^{4}})

\frac{d y}{d x} y = x

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