What is the tangent of 2/(sqrt(x)),\at x= 1/9 ?

The tangent of 2/(sqrt(x)),\at x= 1/9 is y=-27x+9

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tangent of 2/(sqrt(x)),\at x= 1/9

tangent of

\frac{2}{x}, at x = \frac{1}{9}

y = - 27 x + 9

Solution steps

Find the tangent point:

(\frac{1}{9}, 6)

Find the slope of

y = \frac{2}{x} : \frac{d y}{d x} = - \frac{1}{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 = - 27

Find the line with slope m=

- 27

and passing through

(\frac{1}{9}, 6) : y = - 27 x + 9

y = - 27 x + 9

\frac{df}{d t} + f (t) = t + 1

\frac{d}{d x} (x (2))

\frac{d y}{d x} = e^{c s c (6 x) t a n (6 x)}

y^{^{''}} - 3 y^{^{'}} + 8 y = x e^{x}

y^{^{'}} + 9 (tan (9 x)) y = 5 cos (9 x)

What is the tangent of 2/(sqrt(x)),\at x= 1/9 ?
The tangent of 2/(sqrt(x)),\at x= 1/9 is y=-27x+9