\frac{d^2}{dx^2}(2x^3-\sqrt{4-x^2})
\frac{d}{dx}\(2x^3-\sqrt{4-x^2})
Applying sum/difference rule,
(f\pm g\)^'=f^'\pm g^'
=\frac{d}{dx}(2x^3)-\frac{d}{dx}\(\sqrt{4-x^2}
=6x^2 - -\frac{x}{\sqrt{4-x^2}}
Simplifying,
x(6x+\frac{1}{\sqrt{4-x^2}})
=\frac{d}{dx}(x(6x+\frac{1}{\sqrt{4-x^2}})
Applying product rule,
\(f\cdot g)^'=f^'\cdot g+f\cdot g^'
f=x and g=6x+\frac{1}{\sqrt{4-x^2}}
\frac{d}{dx}(x)(6x+\frac{1}{\sqrt{4-x^2}})+\frac{d}{dx}(6x+\frac{1}{\sqrt{4-x^2}})x
d/dx (x) =1
\frac{d}{dx}(6x+\frac{1}{\sqrt{4-x^2}})= 6+\frac{x}{(4-x^2)^{\frac{3}{2}}} {By applying sum/difference rule}
=1(6x+\frac{1}{\sqrt{4-x^2}})+(6+\frac{x}{(4-x^2)^{\frac{3}{2}}})x
Simplifying,
4(3x+\frac{1}{(4-x^2)^{\frac{3}{2}}})