Question 1187769
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \ln^x(x\,+\,8)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln y\ =\ \ln(\ln^x(x\,+\,8)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\,\frac{1}{y}\ =\ \frac{d}{dx}x\,\cdot\,\(\ln\(\ln\(x\,+\,8\)\)\)\ +\ x\,\cdot\,\frac{d}{dx}\ln\(\ln\(x\,+\,8\)\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\,\frac{1}{y}\ =\ \(\ln\(\ln\(x\,+\,8\)\)\)\ +\ \frac{x}{\ln(x\,+\,8)}\,\cdot\,\frac{d}{dx}\ln\(x\,+\,8\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\,\frac{1}{y}\ =\ \(\ln\(\ln\(x\,+\,8\)\)\)\ +\ \frac{x}{\ln(x\,+\,8)}\,\cdot\,\frac{1}{(x\,+\,8)}\,\cdot\,\frac{d}{dx}\(x\,+\,8\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\,\frac{1}{y}\ =\ \(\ln\(\ln\(x\,+\,8\)\)\)\ +\ \frac{x}{\ln(x\,+\,8)}\,\cdot\,\frac{1}{(x\,+\,8)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ y\,\cdot\,\(\ln\(\ln\(x\,+\,8\)\)\)\ +\ \frac{x}{\ln(x\,+\,8)}\,\cdot\,\frac{1}{(x\,+\,8)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \ln^x(x\,+\,8)\,\cdot\,\[\(\ln\(\ln\(x\,+\,8\)\)\)\ +\ \frac{x}{\ln(x\,+\,8)}\,\cdot\,\frac{1}{(x\,+\,8)}\]]


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

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