What is the tangent of sqrt(x+9),\at x=7 ?

The tangent of sqrt(x+9),\at x=7 is y= 1/8 x+25/8

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tangent of sqrt(x+9),\at x=7

tangent of

x + 9, at x = 7

y = \frac{1}{8} x + \frac{25}{8}

Solution steps

Find the tangent point:

(7, 4)

Find the slope of

y = x + 9 : \frac{d y}{d x} = \frac{1}{2 x + 9}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(7, 4) : y = \frac{1}{8} x + \frac{25}{8}

y = \frac{1}{8} x + \frac{25}{8}

\frac{d}{d x} (\frac{2 x - 5}{( x ^{2} + 4 ) ^{3}})

\int x (x + x^{- 1}) d x

\frac{d}{d x} (8 cos (x) sin (x))

\int e^{0.01 x} d x

\frac{d}{d x} (4 8 x)

What is the tangent of sqrt(x+9),\at x=7 ?
The tangent of sqrt(x+9),\at x=7 is y= 1/8 x+25/8