What is the tangent of 5/(sqrt(x)) ?

The tangent of 5/(sqrt(x)) is y=-5/(2a_{0)^{3/2}}x+(15)/(2sqrt(a_{0))}

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tangent of 5/(sqrt(x))

tangent of

\frac{5}{x}

y = - \frac{5}{2 a _{0}^{\frac{3}{2}}} x + \frac{15}{2 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{5}{a _{0}})

Find the slope of

y = \frac{5}{x} : \frac{d y}{d x} = - \frac{5}{2 x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{5}{x}

x = a_{0} : m = - \frac{5}{2 a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{5}{2 a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{5}{a _{0}}) : y = - \frac{5}{2 a _{0}^{\frac{3}{2}}} x + \frac{15}{2 a _{0}}

y = - \frac{5}{2 a _{0}^{\frac{3}{2}}} x + \frac{15}{2 a _{0}}

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

f (x) = x^{2} + 3 x + 4 π, at x = 2.5

\frac{\partial}{\partial y} (2 sin (x) + sin (x + y))

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

\frac{d y}{d x} + 5 y = 6

What is the tangent of 5/(sqrt(x)) ?
The tangent of 5/(sqrt(x)) is y=-5/(2a_{0)^{3/2}}x+(15)/(2sqrt(a_{0))}