What is the tangent of 8/(sqrt(x)) ?

The tangent of 8/(sqrt(x)) is y=-4/(a_{0)^{3/2}}x+(12)/(sqrt(a_{0))}

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tangent of 8/(sqrt(x))

tangent of

\frac{8}{x}

y = - \frac{4}{a _{0}^{\frac{3}{2}}} x + \frac{12}{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{4}{x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{8}{x}

x = a_{0} : m = - \frac{4}{a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{4}{a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{8}{a _{0}}) : y = - \frac{4}{a _{0}^{\frac{3}{2}}} x + \frac{12}{a _{0}}

y = - \frac{4}{a _{0}^{\frac{3}{2}}} x + \frac{12}{a _{0}}

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What is the tangent of 8/(sqrt(x)) ?
The tangent of 8/(sqrt(x)) is y=-4/(a_{0)^{3/2}}x+(12)/(sqrt(a_{0))}