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

The tangent of y=(sqrt(x))/(8x-4),\at x=2 is y=-5/(72sqrt(2))x+(11)/(36sqrt(2))

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tangent of y=(sqrt(x))/(8x-4),\at x=2

tangent of

y = \frac{x}{8 x - 4}, at x = 2

y = - \frac{5}{72 2} x + \frac{11}{36 2}

Solution steps

Find the tangent point:

(2, \frac{1}{6 2})

Find the slope of

y = \frac{x}{8 x - 4} : \frac{d y}{d x} = - \frac{2 x + 1}{8 x ( 2 x - 1 ) ^{2}}

Compute the slope of

y = \frac{x}{8 x - 4}

x = 2 : m = - \frac{5}{72 2}

Find the line with slope m=

- \frac{5}{72 2}

and passing through

(2, \frac{1}{6 2}) : y = - \frac{5}{72 2} x + \frac{11}{36 2}

y = - \frac{5}{72 2} x + \frac{11}{36 2}

x^{4} - 2

- x^{2} + 9 x

\int 3 e^{- x} sin (x) d x

\frac{2 s - 4}{s ^{2} - 4 s + 13}

\frac{d}{d x} (- 0.5 x^{2} + 64 x - 25)

What is the tangent of y=(sqrt(x))/(8x-4),\at x=2 ?
The tangent of y=(sqrt(x))/(8x-4),\at x=2 is y=-5/(72sqrt(2))x+(11)/(36sqrt(2))