What is the tangent of x^2-5 ?

The tangent of x^2-5 is y=2a_{0}x-a_{0}^2-5

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tangent of x^2-5

tangent of

x^{2} - 5

y = 2 a_{0} x - a_{0}^{2} - 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - 5)

Find the slope of

y = x^{2} - 5 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} - 5

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

Find the line with slope m=

2 a_{0}

and passing through

(a_{0}, a_{0}^{2} - 5) : y = 2 a_{0} x - a_{0}^{2} - 5

y = 2 a_{0} x - a_{0}^{2} - 5

x \to 5 lim (500)

x \to 3 + lim (ln (x))

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

\frac{d}{d x} (1 0^{6} x^{2} e^{- x})

\frac{d}{d x} ((4 x + \frac{9}{x})^{2})