# What is the instantaneous rate of change of f(x)=sqrt(x^2+4x+2)  at x=0 ?

May 17, 2016

$\sqrt{2}$

#### Explanation:

The instantaneous rate of change is the value of f'(0).

$f \left(x\right) = \sqrt{{x}^{2} + 4 x + 2} = {\left({x}^{2} + 4 x + 2\right)}^{\frac{1}{2}}$

differentiate using the$\textcolor{b l u e}{\text{ chain rule}}$

$\frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) . g ' \left(x\right) \text{......(A)}$
$\text{-------------------------------------------------------}$

$f \left(g \left(x\right)\right) = {\left({x}^{2} + 4 x + 2\right)}^{\frac{1}{2}}$

$\Rightarrow f ' \left(g \left(x\right)\right) = \frac{1}{2} {\left({x}^{2} + 4 x + 2\right)}^{- \frac{1}{2}}$

and $g \left(x\right) = {x}^{2} + 4 x + 2 \Rightarrow g ' \left(x\right) = 2 x + 4$
$\text{-----------------------------------------------------------}$
Substitute these values into (A)

$\Rightarrow f ' \left(x\right) = \frac{1}{2} {\left({x}^{2} + 4 x + 2\right)}^{- \frac{1}{2}} . \left(2 x + 4\right)$

$= \frac{1}{2} \times \frac{1}{{x}^{2} + 4 x + 2} ^ \left(\frac{1}{2}\right) \times \left(2 x + 4\right)$

$\Rightarrow f ' \left(0\right) = \frac{1}{2} \times \frac{1}{\sqrt{2}} \times 4 = \frac{2}{\sqrt{2}} = \frac{2 \sqrt{2}}{2} = \sqrt{2}$