What is the tangent of y= 1/(sqrt(x))(1.1) ?

The tangent of y= 1/(sqrt(x))(1.1) is y=-(0.55)/(a_{0)^{3/2}}x+(1.65)/(sqrt(a_{0))}

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tangent of y= 1/(sqrt(x))(1.1)

tangent of

y = \frac{1}{x} (1.1)

y = - \frac{0.55}{a _{0}^{\frac{3}{2}}} x + \frac{1.65}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1.1}{a _{0}})

Find the slope of

y = \frac{1}{x} 1.1 : \frac{d y}{d x} = - \frac{0.55}{x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{1}{x} 1.1

x = a_{0} : m = - \frac{0.55}{a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{0.55}{a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{1.1}{a _{0}}) : y = - \frac{0.55}{a _{0}^{\frac{3}{2}}} x + \frac{1.65}{a _{0}}

y = - \frac{0.55}{a _{0}^{\frac{3}{2}}} x + \frac{1.65}{a _{0}}

\int \frac{x}{x ^{2} - 4 x + 4} d x

\int \frac{- 8 x - 9}{x ^{9}} d x

4 x^{3} - 3 x^{2} + 4, at x = 2

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

What is the tangent of y= 1/(sqrt(x))(1.1) ?
The tangent of y= 1/(sqrt(x))(1.1) is y=-(0.55)/(a_{0)^{3/2}}x+(1.65)/(sqrt(a_{0))}