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d/dx [ln(5x^{3} +4)]
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what the chain rule pull out; say if this was just ln(u) ??
okay so it would be...\[\frac{ 5x ^{2} }{ x ^{3} +4}\]
close .. but not quite :)
It should be 3x^{2} at the top
recall\[\frac d{dx}(lnu)=\frac {u'}{u}\] let u = 5x^3 + 4 u' = 15x^2
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you had the right idea tho, and thats good :)
okay, got it! Thanks :)
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