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

1 Answer
Feb 20, 2016

4(3x+\frac{1}{(4-x^2\)^{\frac{3}{2}}})

Explanation:

\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}}})