Confused on this notation partial derivatives

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The discussion clarifies the notation for partial derivatives, particularly the meaning of "d^2" and "dx^2," which indicate second-order derivatives. It explains that "d^2/dt^2" signifies taking the derivative twice, while "d^1" indicates a single derivative. The correct notation for second derivatives is emphasized, showing that \frac{d^2 f}{dx^2} represents the derivative of the first derivative. Misinterpretations of the notation, such as confusing differentials and variables, are cautioned against. Understanding this notation is essential for correctly applying calculus concepts.
mr_coffee
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Hello everyone I have no idea how to start this problem because I'm confused on the notation, what does it mean?
here is a picture:
http://img291.imageshack.us/img291/1177/lastscan2lc.jpg
I know how to take partial derivatives, but the d^2 part is confusing and the dx^2? what the!
 
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The "d^2/dt^2" part means it is a second order derivative. It basically means how many times you take that derivative.

So when you have: \frac{d^1 x^2}{dx^1} then this means... take the derivative of x^2 one time. So you get 2x.

Now if you you had \frac{d^2 x^2}{dx^2} then this means you take the derivative twice. So in TI-89 syntax you would have:

d(d(x^2,x),x)

which equals 2.
 
It may be a but confusing at first, but you'll get used to it. Note that the "square" is at the 'd'-sign in the numerator and above the x (or any other variable) in the denominator. Of course, we still mean the variable x, and not x². In the nominator, it still has to be clear that we're differentiating f, and not f².

So (I'm using normal derivatives here, not partials, but the notation is similar)
\frac{{d^2 f}}{{dx^2 }} = \frac{d}{{dx}}\left( {\frac{{df}}{{dx}}} \right)

But watch out, not one of the following:
\frac{{df^2 }}<br /> {{dx^2 }},\frac{{d^2 f}}<br /> {{d^2 x}},\frac{{df^2 }}<br /> {{d^2 x}}
 
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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