GeorgeOne
GeorgeOne
01-09-2016
Mathematics
contestada
F(x) = 4√(2x³-1)
F'(x) =....?
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nobillionaireNobley
nobillionaireNobley
03-09-2016
[tex]y= \sqrt[4]{(2x^3-1)} \\ y^4=2x^3-1 \\ 2x^3=y^4+1 \\ x^3= \frac{y^4+1}{2} \\ x= \sqrt[3]{\frac{y^4+1}{2}} [/tex]
Therefore, F'(x) = [tex]\sqrt[3]{\frac{y^4+1}{2}}[/tex]
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