Question 1009862
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Since you listed two functions, I choose to differentiate the second one.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ e^{x^3+2x}]


Use the chain rule:


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


Let *[tex \Large u\ =\ x^3\ +\ 2x]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\text{d}y}{\text{d}u}\ =\ e^u\ =\ e^{x^3+2x}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\text{d}u}{\text{d}x}\ =\ 3x^2\ +\ 2]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\text{d}}{\text{d}x}\,e^{x^3+2x}\ =\ \left(3x^2\ +\ 2\ \right)e^{x^3+2x}]


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

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

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