What is the tangent of y= 7/(sqrt(x))(1.7) ?

The tangent of y= 7/(sqrt(x))(1.7) is y=-(5.95)/(a_{0)^{3/2}}x+(17.85)/(sqrt(a_{0))}

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tangent of y= 7/(sqrt(x))(1.7)

tangent of

y = \frac{7}{x} (1.7)

y = - \frac{5.95}{a _{0}^{\frac{3}{2}}} x + \frac{17.85}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{11.9}{a _{0}})

Find the slope of

y = \frac{7}{x} 1.7 : \frac{d y}{d x} = - \frac{5.95}{x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{7}{x} 1.7

x = a_{0} : m = - \frac{5.95}{a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{5.95}{a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{11.9}{a _{0}}) : y = - \frac{5.95}{a _{0}^{\frac{3}{2}}} x + \frac{17.85}{a _{0}}

y = - \frac{5.95}{a _{0}^{\frac{3}{2}}} x + \frac{17.85}{a _{0}}

(\frac{x}{3} + 9)^{\frac{3}{2}}

\int_{0}^{- 1} \frac{x ^{3} + 8}{x + 2} d x

\frac{d y}{d x} + \frac{y}{x} = x^{3} y^{2}

\frac{d}{d x} (ln e^{1 + x^{2}} + 3 x^{4} + cot^{2} (π x))

(\frac{1}{1 + e ^{- x}})^{^{'}}

What is the tangent of y= 7/(sqrt(x))(1.7) ?
The tangent of y= 7/(sqrt(x))(1.7) is y=-(5.95)/(a_{0)^{3/2}}x+(17.85)/(sqrt(a_{0))}