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

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Answer 1 · 2016-02-20T19:06:35

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

#\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}})#

#\(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}}})#