What is the tangent of y= 3/(sqrt(x)),\at x= 1/16 ?

The tangent of y= 3/(sqrt(x)),\at x= 1/16 is y=-96x+18

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tangent of y= 3/(sqrt(x)),\at x= 1/16

tangent of

y = \frac{3}{x}, at x = \frac{1}{16}

y = - 96 x + 18

Solution steps

Find the tangent point:

(\frac{1}{16}, 12)

Find the slope of

y = \frac{3}{x} : \frac{d y}{d x} = - \frac{3}{2 x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{3}{x}

x = \frac{1}{16} : m = - 96

Find the line with slope m=

- 96

and passing through

(\frac{1}{16}, 12) : y = - 96 x + 18

y = - 96 x + 18

(x^{2} + 1) y^{^{'}} = x y

\frac{d}{d x} (ln (tan (x) + sec (x)))

\int cos^{2} (2 u) d u

\int x^{3} - 6 x^{2} + 8 x d x

x \to 0 lim (e^{3 x} - 1)

What is the tangent of y= 3/(sqrt(x)),\at x= 1/16 ?
The tangent of y= 3/(sqrt(x)),\at x= 1/16 is y=-96x+18