What is the tangent of f(x)= 5/(sqrt(x)),\at x= 1/16 ?

The tangent of f(x)= 5/(sqrt(x)),\at x= 1/16 is y=-160x+30

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tangent of f(x)= 5/(sqrt(x)),\at x= 1/16

tangent of

f (x) = \frac{5}{x}, at x = \frac{1}{16}

y = - 160 x + 30

Solution steps

Find the tangent point:

(\frac{1}{16}, 20)

Find the slope of

f (x) = \frac{5}{x} : \frac{df ( x )}{d x} = - \frac{5}{2 x ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{5}{x}

x = \frac{1}{16} : m = - 160

Find the line with slope m=

- 160

and passing through

(\frac{1}{16}, 20) : y = - 160 x + 30

y = - 160 x + 30

\int \frac{( 2 t ^{2} + t t - 1 )}{t ^{2}} d t

\frac{d x}{d t} + 8 x = 0

\frac{d}{d x} (4 x (sin (x) + cos (x)))

\frac{\partial}{\partial x} (x y + b^{x y})

\frac{d y}{d x} = \frac{x - 1}{y ^{3}}

What is the tangent of f(x)= 5/(sqrt(x)),\at x= 1/16 ?
The tangent of f(x)= 5/(sqrt(x)),\at x= 1/16 is y=-160x+30