What is the tangent of y= 5/(sqrt(x))(1.5) ?

The tangent of y= 5/(sqrt(x))(1.5) is y=-(3.75)/(a_{0)^{3/2}}x+(11.25)/(sqrt(a_{0))}

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tangent of y= 5/(sqrt(x))(1.5)

tangent of

y = \frac{5}{x} (1.5)

y = - \frac{3.75}{a _{0}^{\frac{3}{2}}} x + \frac{11.25}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{7.5}{a _{0}})

Find the slope of

y = \frac{5}{x} 1.5 : \frac{d y}{d x} = - \frac{3.75}{x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{5}{x} 1.5

x = a_{0} : m = - \frac{3.75}{a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{3.75}{a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{7.5}{a _{0}}) : y = - \frac{3.75}{a _{0}^{\frac{3}{2}}} x + \frac{11.25}{a _{0}}

y = - \frac{3.75}{a _{0}^{\frac{3}{2}}} x + \frac{11.25}{a _{0}}

y^{^{''}} + 14 y^{^{'}} + 65 y = 0

f (x) = 3 x + 10, (2, 4)

\int \frac{26}{( x - 1 ) ( x ^{2} + 25 )} d x

(x^{2} - 7 x + 4) 25 - x^{2}

\int \frac{x + 1}{3} d x

What is the tangent of y= 5/(sqrt(x))(1.5) ?
The tangent of y= 5/(sqrt(x))(1.5) is y=-(3.75)/(a_{0)^{3/2}}x+(11.25)/(sqrt(a_{0))}