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

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

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

tangent of

f (x) = \frac{2}{x}, at x = 9

y = - \frac{1}{27} x + 1

Solution steps

Find the tangent point:

(9, \frac{2}{3})

Find the slope of

f (x) = \frac{2}{x} : \frac{df ( x )}{d x} = - \frac{1}{x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{2}{x}

x = 9 : m = - \frac{1}{27}

Find the line with slope m=

- \frac{1}{27}

and passing through

(9, \frac{2}{3}) : y = - \frac{1}{27} x + 1

y = - \frac{1}{27} x + 1

\frac{\partial}{\partial z} (- \frac{1}{( y + 2 z ) ^{2}})

\int \frac{3 e ^{2 x}}{1 + e ^{2 x}} d x

f (x) = sin (2 x), 0, \frac{π}{2}

\frac{d}{d x} (5 sin (x) + 3 cos (x))

\frac{d}{d x} (e^{- x y})

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