What is the tangent of f(x)=sqrt(x+5),\at x=4 ?

The tangent of f(x)=sqrt(x+5),\at x=4 is y= 1/6 x+7/3

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tangent of f(x)=sqrt(x+5),\at x=4

tangent of

f (x) = x + 5, at x = 4

y = \frac{1}{6} x + \frac{7}{3}

Solution steps

Find the tangent point:

(4, 3)

Find the slope of

f (x) = x + 5 : \frac{df ( x )}{d x} = \frac{1}{2 x + 5}

Compute the slope of

f (x) = x + 5

x = 4 : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(4, 3) : y = \frac{1}{6} x + \frac{7}{3}

y = \frac{1}{6} x + \frac{7}{3}

(2 x - y) d x + (2 y - x) d y = 0

\frac{d}{d x} ((1 - x)^{- 2})

y^{^{'}} = \frac{y}{x} + 1

y^{^{'''}} + 2 y^{^{''}} - 5 y^{^{'}} - 6 y = 0

\int_{0}^{π} 16 sin (x) d x

What is the tangent of f(x)=sqrt(x+5),\at x=4 ?
The tangent of f(x)=sqrt(x+5),\at x=4 is y= 1/6 x+7/3