Question 375065
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You get one question per post.  Since the first one is trivial, I'll do the second.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ x^x]


Take the natural log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(y)\ =\ \ln\left(x^x\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(y)\ =\ x\ln(x)]


Differentiate both sides with respect to *[tex \Large x] using the Chain Rule on the left and the Product Rule on the right:


Let *[tex \Large u\ =\ \ln(y)].


Then *[tex \Large \frac{du}{dy}\ =\ \frac{1}{y}]


Hence:


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


Multiply both sides by *[tex \Large y\ =\ x^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ x^x\left(1\ +\ \ln(x)\right)]




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