What is the tangent of 1/(2sqrt(x)),\at x=4 ?

The tangent of 1/(2sqrt(x)),\at x=4 is y=-1/32 x+3/8

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tangent of 1/(2sqrt(x)),\at x=4

tangent of

\frac{1}{2 x}, at x = 4

y = - \frac{1}{32} x + \frac{3}{8}

Solution steps

Find the tangent point:

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

Find the slope of

y = \frac{1}{2 x} : \frac{d y}{d x} = - \frac{1}{4 x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{1}{2 x}

x = 4 : m = - \frac{1}{32}

Find the line with slope m=

- \frac{1}{32}

and passing through

(4, \frac{1}{4}) : y = - \frac{1}{32} x + \frac{3}{8}

y = - \frac{1}{32} x + \frac{3}{8}

x \to 8 + lim (x - 8)

\frac{d x}{d t} = x^{2} - 2 x + 2

sin (2 t) cos (3 t)

\int (2 x - 1) y^{2} sin (3 x y) d x

\int 15 (tan (x)) (ln (cos (x))) d x

What is the tangent of 1/(2sqrt(x)),\at x=4 ?
The tangent of 1/(2sqrt(x)),\at x=4 is y=-1/32 x+3/8