[ln(u)]' = u' / u
So, [ln(2x2+x+3)]' = (2x2+x+3)' / (2x2+x+3) = (4x+1) / (2x2+x+3)
Maria S.
asked 06/05/21What is h′(x) when h(x)=ln(2x^2+x+3)?
[ln(u)]' = u' / u
So, [ln(2x2+x+3)]' = (2x2+x+3)' / (2x2+x+3) = (4x+1) / (2x2+x+3)
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