What is the tangent of 8x^{-2} ?

The tangent of 8x^{-2} is y=-(16)/(a_{0)^3}x+(24)/(a_{0)^2}

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tangent of 8x^{-2}

tangent of

8 x^{- 2}

y = - \frac{16}{a _{0}^{3}} x + \frac{24}{a _{0}^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{8}{a _{0}^{2}})

Find the slope of

y = 8 x^{- 2} : \frac{d y}{d x} = - \frac{16}{x ^{3}}

Compute the slope of

y = 8 x^{- 2}

x = a_{0} : m = - \frac{16}{a _{0}^{3}}

Find the line with slope m=

- \frac{16}{a _{0}^{3}}

and passing through

(a_{0}, \frac{8}{a _{0}^{2}}) : y = - \frac{16}{a _{0}^{3}} x + \frac{24}{a _{0}^{2}}

y = - \frac{16}{a _{0}^{3}} x + \frac{24}{a _{0}^{2}}

\int - x d x

x \to 0 lim (arccot (\frac{1}{x}))

\frac{d}{d x} ((- x^{2} + 2 x) (3 x + 2) (x - 1))

f (x) = 3 x^{2}, (1, 3)

y^{^{'}} = \frac{e ^{x}}{y}

What is the tangent of 8x^{-2} ?
The tangent of 8x^{-2} is y=-(16)/(a_{0)^3}x+(24)/(a_{0)^2}