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}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

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))