What is the tangent of f(x)= 1/(sqrt(8x)),\at x=9 ?

The tangent of f(x)= 1/(sqrt(8x)),\at x=9 is y=-1/(108sqrt(2))x+1/(4sqrt(2))

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tangent of f(x)= 1/(sqrt(8x)),\at x=9

tangent of

f (x) = \frac{1}{8 x}, at x = 9

y = - \frac{1}{108 2} x + \frac{1}{4 2}

Solution steps

Find the tangent point:

(9, \frac{1}{6 2})

Find the slope of

f (x) = \frac{1}{8 x} : \frac{df ( x )}{d x} = - \frac{1}{4 2 x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{1}{8 x}

x = 9 : m = - \frac{1}{108 2}

Find the line with slope m=

- \frac{1}{108 2}

and passing through

(9, \frac{1}{6 2}) : y = - \frac{1}{108 2} x + \frac{1}{4 2}

y = - \frac{1}{108 2} x + \frac{1}{4 2}

\frac{\partial}{\partial x} (2 x^{3} y^{2} + 2 y + 4 x)

\int x^{3} x^{4} + 1 d x

\frac{d ^{2} y}{d x ^{2}} + a^{2} \cdot y - e^{b - x} - 1 = 0

\frac{1}{y} d x - \frac{x}{( ln ( x ) ) ^{3}} d y = 0

\frac{1}{y ^{2}} - \frac{3}{y ^{4}} (y + 5 y^{3})

What is the tangent of f(x)= 1/(sqrt(8x)),\at x=9 ?
The tangent of f(x)= 1/(sqrt(8x)),\at x=9 is y=-1/(108sqrt(2))x+1/(4sqrt(2))