Question 1187766
Sunday, April 2, 2017
7:06 PM

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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \(6\,+\,3x\)^{5/x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(y)\ =\ \ln\(\(6\,+\,3x\)^{5/x}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(y)\ =\ \frac{5}{x}\,\ln\(6\,+\,3x\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\cdot\frac{1}{y}\ =\ \frac{d}{dx}\,\[\frac{5}{x}\,\ln\(6\,+\,3x\)\]]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\cdot\frac{1}{y}\ =\ \(\frac{d}{dx}\,\frac{5}{x}\)\(\ln\(6\,+\,3x\)\)\ +\ \(\frac{5}{x}\)\(\frac{d}{dx}\,\ln\(6\,+\,3x\)\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\cdot\frac{1}{y}\ =\ \(-\frac{5}{x^2}\)\(\ln\(6\,+\,3x\)\)\ +\ \(\frac{5}{x}\)\(\frac{1}{6\,+\,3x}\)\,\frac{d}{dx}(6\,+\,3x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\cdot\frac{1}{y}\ =\ \(-\frac{5}{x^2}\)\(\ln\(6\,+\,3x\)\)\ +\ \(\frac{15}{x}\)\(\frac{1}{6\,+\,3x}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ \(6\,+\,3x\)^{5/x}\[\(-\frac{5}{x^2}\)\(\ln\(6\,+ \,3x\)\)\ +\ \(\frac{15}{x}\)\(\frac{1}{6\,+\,3x}\)\]]


For part b, evaluate *[tex \Large y'(1)]


For part c,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y(1)\ =\ y'(1)(x\,-\,1)]


You can do your own arithmetic.




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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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