What is the tangent of 1/(2sqrt(x+2)) ?

The tangent of 1/(2sqrt(x+2)) is y=-1/(4(a_{0)+2)^{3/2}}x+(3a_{0}+4)/(4(a_{0)+2)^{3/2}}

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tangent of 1/(2sqrt(x+2))

tangent of

\frac{1}{2 x + 2}

y = - \frac{1}{4 ( a _{0} + 2 ) ^{\frac{3}{2}}} x + \frac{3 a _{0} + 4}{4 ( a _{0} + 2 ) ^{\frac{3}{2}}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{2 a _{0} + 2})

Find the slope of

y = \frac{1}{2 x + 2} : \frac{d y}{d x} = - \frac{1}{4 ( x + 2 ) ^{\frac{3}{2}}}

Compute the slope of

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

x = a_{0} : m = - \frac{1}{4 ( a _{0} + 2 ) ^{\frac{3}{2}}}

Find the line with slope m=

- \frac{1}{4 ( a _{0} + 2 ) ^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{1}{2 a _{0} + 2}) : y = - \frac{1}{4 ( a _{0} + 2 ) ^{\frac{3}{2}}} x + \frac{3 a _{0} + 4}{4 ( a _{0} + 2 ) ^{\frac{3}{2}}}

y = - \frac{1}{4 ( a _{0} + 2 ) ^{\frac{3}{2}}} x + \frac{3 a _{0} + 4}{4 ( a _{0} + 2 ) ^{\frac{3}{2}}}

\int x e^{9 x^{2}} d x

\frac{d}{d x} (\frac{48}{x})

\frac{d}{d x} (e^{- x} + e^{2 x})

f (x) = 5 x - 4 x^{2}, (- 3, - 51)

\int_{0}^{1} x^{n} d x

What is the tangent of 1/(2sqrt(x+2)) ?
The tangent of 1/(2sqrt(x+2)) is y=-1/(4(a_{0)+2)^{3/2}}x+(3a_{0}+4)/(4(a_{0)+2)^{3/2}}