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

The tangent of sqrt(x),\at x=9 is y= 1/6 x+3/2

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

tangent of

x, at x = 9

y = \frac{1}{6} x + \frac{3}{2}

Solution steps

Find the tangent point:

(9, 3)

Find the slope of

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

Compute the slope of

y = x

x = 9 : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(9, 3) : y = \frac{1}{6} x + \frac{3}{2}

y = \frac{1}{6} x + \frac{3}{2}

\frac{\partial}{\partial y} (4 y)

(10, 5), (1, 12)

\frac{1}{s ^{3} + 1}

x \to 0 lim (x \cdot e^{- \frac{1}{x}})

\int \frac{x ^{2}}{( x + 1 ) ^{4}} d x

What is the tangent of sqrt(x),\at x=9 ?
The tangent of sqrt(x),\at x=9 is y= 1/6 x+3/2