What is the tangent of 1/(2sqrt(x)),\at x=64 ?

The tangent of 1/(2sqrt(x)),\at x=64 is y=-1/2048 x+3/32

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tangent of 1/(2sqrt(x)),\at x=64

tangent of

\frac{1}{2 x}, at x = 64

y = - \frac{1}{2048} x + \frac{3}{32}

Solution steps

Find the tangent point:

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

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 = 64 : m = - \frac{1}{2048}

Find the line with slope m=

- \frac{1}{2048}

and passing through

(64, \frac{1}{16}) : y = - \frac{1}{2048} x + \frac{3}{32}

y = - \frac{1}{2048} x + \frac{3}{32}

\int \frac{1}{x 121 x ^{2} - 121} d x

\int y + x d x

x \to 3 lim (\frac{∣ 3 - x ∣}{x - 3})

(2 x^{3} - 3 x + 1)^{^{'}}

y = (x^{2} - 2 x + 7) 25 - x^{2}

What is the tangent of 1/(2sqrt(x)),\at x=64 ?
The tangent of 1/(2sqrt(x)),\at x=64 is y=-1/2048 x+3/32