What is the tangent of f(x)= 2/(sqrt(x)),\at x= 1/16 ?

The tangent of f(x)= 2/(sqrt(x)),\at x= 1/16 is y=-64x+12

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tangent of f(x)= 2/(sqrt(x)),\at x= 1/16

tangent of

f (x) = \frac{2}{x}, at x = \frac{1}{16}

y = - 64 x + 12

Solution steps

Find the tangent point:

(\frac{1}{16}, 8)

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 = \frac{1}{16} : m = - 64

Find the line with slope m=

- 64

and passing through

(\frac{1}{16}, 8) : y = - 64 x + 12

y = - 64 x + 12

\int \frac{1}{x \cdot ( ln ( x ) ) ^{p}} d x

\frac{d y}{d x} = - 11

f (x) = 3 x^{2} - x + 5

x \to - 2 - lim (\frac{1 - x}{x ^{2} - 4})

n = 1 \sum \infty 3^{n} x^{n} n!

What is the tangent of f(x)= 2/(sqrt(x)),\at x= 1/16 ?
The tangent of f(x)= 2/(sqrt(x)),\at x= 1/16 is y=-64x+12