What is the tangent of 2sqrt(x),\at x=9 ?

The tangent of 2sqrt(x),\at x=9 is y= 1/3 x+3

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tangent of 2sqrt(x),\at x=9

tangent of

2 x, at x = 9

y = \frac{1}{3} x + 3

Solution steps

Find the tangent point:

(9, 6)

Find the slope of

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

Compute the slope of

y = 2 x

x = 9 : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(9, 6) : y = \frac{1}{3} x + 3

y = \frac{1}{3} x + 3

f (x) = \frac{8 x}{x ^{2} - 1}, at x = 3

(\frac{6 x ^{2}}{8 - x})^{3}

x^{- 5}

\int 4 sin (5 x) d x

\int \frac{1}{x ^{\frac{2}{3}} ( 8 + x ^{\frac{1}{3}} )} d x

What is the tangent of 2sqrt(x),\at x=9 ?
The tangent of 2sqrt(x),\at x=9 is y= 1/3 x+3