Question 1164751
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \(4x^3\ +\ \frac{1}{2}x^2\ +\ 9e^0\)^3]


First thing, *[tex \Large e^0\ =\ 1], so: 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \(4x^3\ +\ \frac{1}{2}x^2\ +\ 9\)^3]


Use the Chain rule:  *[tex \Large \frac{dy}{dx}\ =\ \frac{dy}{du}\,\cdot\,\frac{du}{dx}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ 3\(4x^3\ +\ \frac{1}{2}x^2\ +\ 9\)^2\,\cdot\,\(12x^2\ +\ x\)]

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