What is the tangent of 8/x ?

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

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tangent of 8/x

tangent of

\frac{8}{x}

y = - \frac{8}{a _{0}^{2}} x + \frac{16}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = \frac{8}{x}

x = a_{0} : m = - \frac{8}{a _{0}^{2}}

Find the line with slope m=

- \frac{8}{a _{0}^{2}}

and passing through

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

y = - \frac{8}{a _{0}^{2}} x + \frac{16}{a _{0}}

\frac{d x}{d t} + 5 x = 50

\frac{d}{d x} (\frac{7 e ^{x}}{2 e ^{x} + 1})

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

\frac{x ^{3} - x}{1 + x ^{2}}

f (x) = x^{2} - x, (- 4, 20)