What is the tangent of y=sqrt(x),(64,8) ?

The tangent of y=sqrt(x),(64,8) is y= 1/16 x+4

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tangent of y=sqrt(x),(64,8)

tangent of

y = x, (64, 8)

y = \frac{1}{16} x + 4

Solution steps

Find the slope of

y = x : \frac{d y}{d x} = \frac{1}{2 x}

Compute the slope of

y = x

(64, 8) : m = \frac{1}{16}

Find the line with slope m=

\frac{1}{16}

and passing through

(64, 8) : y = \frac{1}{16} x + 4

y = \frac{1}{16} x + 4

\frac{12 s}{12 s - 30 s ^{2}}

\frac{d y}{d x} = \frac{( e ^{y} )}{x y}

\frac{1 - 5 t}{4 + t}

\int sin (8 x) cos (8 x) d x

(x - 1)^{2}

What is the tangent of y=sqrt(x),(64,8) ?
The tangent of y=sqrt(x),(64,8) is y= 1/16 x+4