Question 983313
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 3x\ +\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\text{d}u}{\text{d}x}\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ u^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\text{d}y}{\text{d}u}\ =\ 2u\ =\ 6x\ +\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\text{d}y}{\text{d}x}\ =\ \frac{\text{d}y}{\text{d}u}\frac{\text{d}u}{\text{d}x}\ =\ (6x\ +\ 2)*3\ =\ 18x\ +\ 6]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \