What is the tangent of 3sqrt(x),\at x=16 ?

The tangent of 3sqrt(x),\at x=16 is y= 3/8 x+6

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tangent of 3sqrt(x),\at x=16

tangent of

3 x, at x = 16

y = \frac{3}{8} x + 6

Solution steps

Find the tangent point:

(16, 12)

Find the slope of

y = 3 x : \frac{d y}{d x} = \frac{3}{2 x}

Compute the slope of

y = 3 x

x = 16 : m = \frac{3}{8}

Find the line with slope m=

\frac{3}{8}

and passing through

(16, 12) : y = \frac{3}{8} x + 6

y = \frac{3}{8} x + 6

\int x \cdot sin (2 \cdot x) d x

450 (1 + \frac{3 t}{29 + t ^{2}}), 4

x \to 0 lim (\frac{( x + 4 )}{2})

y = 2 x e^{- x}

x \to \infty lim (2.5)

What is the tangent of 3sqrt(x),\at x=16 ?
The tangent of 3sqrt(x),\at x=16 is y= 3/8 x+6