Question 1024223
<font face="Times New Roman" size="+2">


Product Rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\text{d}}{\text{d}x}(u\ \cdot\ v)\ =\ u\ \cdot\ \frac{\text{d}v}{\text{d}x}\ +\ v\ \cdot\ \frac{\text{d}u}{\text{d}x}]


Let *[tex \Large u\ =\ x\ +\ 1], then *[tex \Large \frac{\text{d}u}{\text{d}x}\ =\ 1]


Let *[tex \Large v\ =\ 9\sqrt{x}\ +\ 8], then *[tex \Large \frac{\text{d}v}{\text{d}x}\ =\ \frac{9}{2\sqrt{x}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\text{d}}{\text{d}x}(x\ +\ 1)(9\sqrt{x}\ +\ 8)\ =\ \frac{9x\ +\ 9}{2\sqrt{x}}\ +\ 9\sqrt{x}\ +\ 8\ =\ \frac{27x\ +\ 16\sqrt{x}\ +\ 9}{2\sqrt{x}}]


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

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

</font>