What is the tangent of 2/(sqrt(x+5)),\at x=4 ?

The tangent of 2/(sqrt(x+5)),\at x=4 is y=-1/27 x+22/27

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tangent of 2/(sqrt(x+5)),\at x=4

tangent of

\frac{2}{x + 5}, at x = 4

y = - \frac{1}{27} x + \frac{22}{27}

Solution steps

Find the tangent point:

(4, \frac{2}{3})

Find the slope of

y = \frac{2}{x + 5} : \frac{d y}{d x} = - \frac{1}{( x + 5 ) ^{\frac{3}{2}}}

Compute the slope of

y = \frac{2}{x + 5}

x = 4 : m = - \frac{1}{27}

Find the line with slope m=

- \frac{1}{27}

and passing through

(4, \frac{2}{3}) : y = - \frac{1}{27} x + \frac{22}{27}

y = - \frac{1}{27} x + \frac{22}{27}

f (x) = 2 x^{2} - x - 3, at x = 1

y = 2 - 6 x

y = \frac{2 x + 4}{1 - x}

3 y^{^{''}} + 2 y^{^{'}} = 0

\frac{d N}{d t} + N = Nt e^{t + 4}

What is the tangent of 2/(sqrt(x+5)),\at x=4 ?
The tangent of 2/(sqrt(x+5)),\at x=4 is y=-1/27 x+22/27