What is the tangent of x^2-3 ?

The tangent of x^2-3 is y=2a_{0}x-a_{0}^2-3

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tangent of x^2-3

tangent of

x^{2} - 3

y = 2 a_{0} x - a_{0}^{2} - 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 3)

Find the slope of

y = x^{2} - 3 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} - 3

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

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} - 3) : y = 2 a_{0} x - a_{0}^{2} - 3

y = 2 a_{0} x - a_{0}^{2} - 3

f (x) = x^{2} - cos (2 x)

\frac{d}{d x} (ln ((x^{2} + 1)^{2}))

y = x^{2} - 5, (2, - 1)

\frac{\partial}{\partial x} (x \cdot ln (2 y))