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

The tangent of f(x)= 1/(sqrt(x)),\at x= 1/16 is y=-32x+6

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

tangent of

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

y = - 32 x + 6

Solution steps

Find the tangent point:

(\frac{1}{16}, 4)

Find the slope of

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

Compute the slope of

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

x = \frac{1}{16} : m = - 32

Find the line with slope m=

- 32

and passing through

(\frac{1}{16}, 4) : y = - 32 x + 6

y = - 32 x + 6

x^{2} + 4 x - 5

\frac{d y}{d x} = \frac{( x - 1 )}{( y - 1 )}

\int_{0}^{0.1} x arctan (4 x) d x

\frac{d y}{d t} = \frac{8 t y}{t ^{2} + 1}

y^{^{'}} = 1 - t + y^{2} - t y^{2}

What is the tangent of f(x)= 1/(sqrt(x)),\at x= 1/16 ?
The tangent of f(x)= 1/(sqrt(x)),\at x= 1/16 is y=-32x+6