December 2001:
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12/31/01) Find the result of dividing the 65th derivative of 2x66 by 65 factorial.
When taking the derivative of an exponential function, we bring down the exponent as a coefficient and subtract one from the exponent for the new one. Therefore, the 65th derivative of 2x66 is simply 2*66!x, which turns into 2*66x or 132x when divided by 65!. A little thinking would be sufficient, right?
______________________________________ 12/24/01)
Compute:
Topic:
Derivatives
Solution:
12/24/01) First let us simplify the expression in the limit. Let's call it A. Now wouldn't you agree that A=(b+A)1/2, since it is the same expression A in the square root? Therefore, by squaring both sides and solving for A, we would get A= 1±(1-4b)1/2/2, of which the negative part can be eliminated. When b approaches zero, the expression is just (1+1)/2, which is 1. Isn't it nice?
______________________________________ 12/17/01) Where does the function y = |3x2| - |x4| -9 change its concavity?
Topic:
Limits and Algebra
Where does y = | |3x2| - |x4| | change from decreasing to increasing or vice versa?Solution:
The first part is very simple, since the absolute values are meaningless. 3x2 and x4 are always positive. Therefore, by taking the second derivative of the function, we would find the inflection points to be -root2/2 and +root2/2.
The second part is more complex. Firstly, the inner absolute values can be eliminated again. So we only work with |x2-x4|. Since x2 and x4 meet at points x=1 and x=-1, those are the places the function is not differentiable at, but should be analyzed. This function can be rewritten as x2 - x4 when x is between -1 and 1; and x4 - x2 when x is anywhere else. Therefore the first derivative of the function is 2x-4x3 between -1 and 1. Hence, it is 4x3-2x anywhere else. Setting these equal to zero, we get x=0, x=-root2/2, x=root2/2. However, the points -1 and 1, at which the function is not differentiable, are also extremas, since the function is flipped at the points. So the answer set is {-1,-root2/2,0,root2/2,1}. Although tedious, it was lovely, wasn't it?
Topic:
Extremas and Absolute Values
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12/10/01) How many local max/mins are there in the function y = sin (1/x), between -2/π and 2/π one of the local extremas? What about 1/1000π?
The function sin(1/x) oscillates infinitely, when x approaches zero. When differentiated, the derivative function is -cos(1/x)/x2. The derivative function has infinite zeros. Therefore, the number of local max/mins are also infinite. Wherever the cos(1/x) equals zero, the original function has local extremas. The formula for the extremas is 2/(nπ) where n can be any nonzero odd integer. Therefore, 2/2001π is an extrema, whereas 1/1000π is not. Fascinating, huh?
Topic:
Extremas
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12/03/01)
The sin and cos function are cyclical in their derivatives. In every four interval, they start to repeat their pattern. So the answer to part 1 is simply the third derivative of the given function, which is sin x - cos x. By the way, the csc and sec can be easily simplified to sines and cosines.
For the second part, you have to be a little cleverer! x100 becomes a coefficient times x in the 99th derivative. Therefore, in the hundreth derivative, it is just zero and for the solution of this problem, it can be ignored!! The rest is just like the first part, since it is cyclical. 110th derivative is just like the second one. So the answer is -10cos(x)+8sin(x). GREAT, huh?
Topic:
Cyclical Derivatives
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