Question 1189661
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{1}{\sqrt{2x\,-\,3}}\ =\ \(2x\,-\,3\)^{\small{-\frac{1}{2}}}]


*[tex \Large f(x)] is continuous only on the interval *[tex \Large \(\frac{3}{2},\infty\)] so the derivative is defined only on that interval.


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ -\frac{1}{2}\cdot\(2x\,-\,3\)^{\small{-\frac{3}{2}}}\cdot 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dx}\ =\ -\frac{1}{(2x\,-\,3\)^{\small{\frac{3}{2}}}}]

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

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