ln Diff Example - # 3

[Jason76]

\(\displaystyle y = (x^{3} + 2)^{2} (x^{5} + 4)^{4}\)

\(\displaystyle \ln(y) = \ln[(x^{3} + 2)^{2} (x^{5} + 4)^{4}]\)

\(\displaystyle \ln(y) = \ln[(x^{3} + 2)^{2}] + \ln [(x^{5} + 4)^{4}]\)

\(\displaystyle \ln(y) = 2 \ln [(x^{3} + 2)] + 4 \ln [(x^{5} + 4)]\)

 

[Deleted member 4993]

\(\displaystyle y = (x^{3} + 2)^{2} (x^{5} + 4)^{4}\)

\(\displaystyle \ln(y) = \ln[(x^{3} + 2)^{2} (x^{5} + 4)^{4}]\) .............. why ln? ...............use product rule and chain rule.

\(\displaystyle \ln(y) = \ln[(x^{3} + 2)^{2}] + \ln [(x^{5} + 4)^{4}]\)

\(\displaystyle \ln(y) = 2 \ln [(x^{3} + 2)] + 4 \ln [(x^{5} + 4)]\)

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[HallsofIvy]

Subhotosh Kahn, I suspect that these were exercises in using logarthmic differentation, even when it is not, strictly speaking, necessary.

Jason76, why are you taking the logarithm then NOT differentiating?

Yes, \(\displaystyle ln(y)= ln\left((x^3+ 2)^2(x^5+ 4)^4\right)= 2ln(x^3+ 2)+ 4ln(x^5+ 4)\)

Now \(\displaystyle (ln(y))'= \dfrac{y'}{y}\), \(\displaystyle (ln(x^3+ 2))'= \dfrac{3x^2}{x^3+ 2}\), and \(\displaystyle (ln(x^5+ 4))'= \dfrac{5x^4}{x^5+ 4}\).

Now can you put those together (don't forget the factors of "2" and "4")?