What is the tangent of sqrt(x),\at x=16 ?

The tangent of sqrt(x),\at x=16 is y= 1/8 x+2

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tangent of sqrt(x),\at x=16

tangent of

x, at x = 16

y = \frac{1}{8} x + 2

Solution steps

Find the tangent point:

(16, 4)

Find the slope of

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

Compute the slope of

y = x

x = 16 : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(16, 4) : y = \frac{1}{8} x + 2

y = \frac{1}{8} x + 2

\int_{2}^{x} (3 t^{2} - 1) d t

F (r) = \frac{5}{r ^{3}}

\frac{d}{d x} (\frac{3}{x + 7})

\int \frac{csc ^{4} ( 2 x )}{cot ^{2} ( 2 x )} d x

What is the tangent of sqrt(x),\at x=16 ?
The tangent of sqrt(x),\at x=16 is y= 1/8 x+2