Solving a Simple Derivative Problem: Understanding the Disappearance of 3x^1/2

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The discussion centers on a confusion regarding the derivative of the function involving 3x^1/2. The original poster struggles with understanding why this term seems to disappear in the textbook's method. A key point highlighted is the conversion of 1/√x to x^(-1/2), which is essential for applying the product rule correctly. The correct differentiation involves applying the product rule to both parts of the function, leading to clarity on the disappearance of the term. This explanation aims to resolve the misunderstanding and clarify the differentiation process.
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Here is the equation and my attempt on a dry-erase board:

bijd3b.jpg


My steps are similar to the textbook's up until I hit that 3x^1/2. Why is it disappearing with their method? It's late at night and my brain is fried; I have a feeling this will be painfully obvious to me in the morning, but in case it isn't, perhaps someone can fill me in on what I missed?
 
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DinosaurEgg said:
Here is the equation and my attempt on a dry-erase board:

bijd3b.jpg


My steps are similar to the textbook's up until I hit that 3x^1/2. Why is it disappearing with their method? It's late at night and my brain is fried; I have a feeling this will be painfully obvious to me in the morning, but in case it isn't, perhaps someone can fill me in on what I missed?

Recall that ##1/\sqrt{x} = x^{-1/2},## so
\frac{d}{dx} \frac{2x-1}{\sqrt{x}} = x^{-1/2}\frac{d}{dx} (2x-1)<br /> + (2x-1) \frac{d}{dx} x^{-1/2}.
You had written ##(d/dx) x^{1/2}.##

RGV
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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