What is the tangent of f(x)=sqrt(x+3),\at x=13 ?

The tangent of f(x)=sqrt(x+3),\at x=13 is y= 1/8 x+19/8

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tangent of f(x)=sqrt(x+3),\at x=13

tangent of

f (x) = x + 3, at x = 13

y = \frac{1}{8} x + \frac{19}{8}

Solution steps

Find the tangent point:

(13, 4)

Find the slope of

f (x) = x + 3 : \frac{df ( x )}{d x} = \frac{1}{2 x + 3}

Compute the slope of

f (x) = x + 3

x = 13 : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(13, 4) : y = \frac{1}{8} x + \frac{19}{8}

y = \frac{1}{8} x + \frac{19}{8}

\frac{d}{d x} (x^{2} + π^{2} + x^{π})

x \to \infty lim (24)

6 y^{^{'}} + 7 y = 5

\frac{d ^{2}}{d x ^{2}} (sin (x) - cos (x))

6 (x^{2} + y^{2}) d x + 5 x y d y = 0

What is the tangent of f(x)=sqrt(x+3),\at x=13 ?
The tangent of f(x)=sqrt(x+3),\at x=13 is y= 1/8 x+19/8