What is the tangent of y= 6/(sqrt(x)) ?

The tangent of y= 6/(sqrt(x)) is y=-3/(a_{0)^{3/2}}x+9/(sqrt(a_{0))}

tangent of y= 6/(sqrt(x))

tangent of

y = \frac{6}{x}

y = - \frac{3}{a _{0}^{\frac{3}{2}}} x + \frac{9}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{6}{a _{0}})

Find the slope of

y = \frac{6}{x} : \frac{d y}{d x} = - \frac{3}{x ^{\frac{3}{2}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{3}{a _{0}^{\frac{3}{2}}}

Find the line with slope m=

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

and passing through

(a_{0}, \frac{6}{a _{0}}) : y = - \frac{3}{a _{0}^{\frac{3}{2}}} x + \frac{9}{a _{0}}

y = - \frac{3}{a _{0}^{\frac{3}{2}}} x + \frac{9}{a _{0}}

\int 7 (sin (x))^{4} (cos (x))^{2} d x

x \frac{d y}{d x} = x^{3} + 14 x^{3} y

\int c x y + \frac{1}{4}

t an g e n t f (x) = \frac{x}{4 x - 7}, at x = 1

t an g e n t - 5 x^{2} + x - 2

What is the tangent of y= 6/(sqrt(x)) ?
The tangent of y= 6/(sqrt(x)) is y=-3/(a_{0)^{3/2}}x+9/(sqrt(a_{0))}