# What is the second derivative of f(x)= 2x^3- sqrt(4-x^2)?

Feb 20, 2016

$4 \left(3 x + \setminus \frac{1}{{\left(4 - {x}^{2} \setminus\right)}^{\setminus \frac{3}{2}}}\right)$

#### Explanation:

$\setminus \frac{{d}^{2}}{{\mathrm{dx}}^{2}} \left(2 {x}^{3} - \setminus \sqrt{4 - {x}^{2}}\right)$

$\setminus \frac{d}{\mathrm{dx}} \setminus \left(2 {x}^{3} - \setminus \sqrt{4 - {x}^{2}}\right)$

Applying sum/difference rule,
${\left(f \setminus \pm g \setminus\right)}^{'} = {f}^{'} \setminus \pm {g}^{'}$

=\frac{d}{dx}(2x^3)-\frac{d}{dx}\(\sqrt{4-x^2}

$= 6 {x}^{2}$ - $- \setminus \frac{x}{\setminus \sqrt{4 - {x}^{2}}}$

Simplifying,
$x \left(6 x + \setminus \frac{1}{\setminus \sqrt{4 - {x}^{2}}}\right)$

=\frac{d}{dx}(x(6x+\frac{1}{\sqrt{4-x^2}})

Applying product rule,

$\setminus {\left(f \setminus \cdot g\right)}^{'} = {f}^{'} \setminus \cdot g + f \setminus \cdot {g}^{'}$

$f = x$ and $g = 6 x + \setminus \frac{1}{\setminus \sqrt{4 - {x}^{2}}}$

$\setminus \frac{d}{\mathrm{dx}} \left(x\right) \left(6 x + \setminus \frac{1}{\setminus \sqrt{4 - {x}^{2}}}\right) + \setminus \frac{d}{\mathrm{dx}} \left(6 x + \setminus \frac{1}{\setminus \sqrt{4 - {x}^{2}}}\right) x$

$\frac{d}{\mathrm{dx}} \left(x\right) = 1$

$\setminus \frac{d}{\mathrm{dx}} \left(6 x + \setminus \frac{1}{\setminus \sqrt{4 - {x}^{2}}}\right)$= $6 + \setminus \frac{x}{{\left(4 - {x}^{2}\right)}^{\setminus \frac{3}{2}}}$ {By applying sum/difference rule}

$= 1 \left(6 x + \setminus \frac{1}{\setminus \sqrt{4 - {x}^{2}}}\right) + \left(6 + \setminus \frac{x}{{\left(4 - {x}^{2}\right)}^{\setminus \frac{3}{2}}}\right) x$

Simplifying,

$4 \left(3 x + \setminus \frac{1}{{\left(4 - {x}^{2}\right)}^{\setminus \frac{3}{2}}}\right)$