What is the tangent of y=sqrt(x+5),\at x=11 ?

The tangent of y=sqrt(x+5),\at x=11 is y= 1/8 x+21/8

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tangent of y=sqrt(x+5),\at x=11

tangent of

y = x + 5, at x = 11

y = \frac{1}{8} x + \frac{21}{8}

Solution steps

Find the tangent point:

(11, 4)

Find the slope of

y = x + 5 : \frac{d y}{d x} = \frac{1}{2 x + 5}

Compute the slope of

y = x + 5

x = 11 : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(11, 4) : y = \frac{1}{8} x + \frac{21}{8}

y = \frac{1}{8} x + \frac{21}{8}

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

\int_{- 1}^{4} x - 2 x^{2} d x

\frac{\partial}{\partial x} (x y e^{y})

f (x) = 2^{s i n (x)}

y = (\frac{8}{x})^{5 x}

What is the tangent of y=sqrt(x+5),\at x=11 ?
The tangent of y=sqrt(x+5),\at x=11 is y= 1/8 x+21/8