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Calculus1
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Suppose all the derivatives of some function f exist at 0 and that the Taylor series for f about x = 0 is given by: x + x^2/2 + x^3/3 + x^4/4 + . . . Give the value of each of the following. f'(0): f"(0): f"'(0): f^10(0):
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use the derivative each time and then set x = 0 to get the answer but gor f^10(0) i have no idea how to do it
f'(0):1 f"(0):1 f"'(0):2
for f^10(0) i got zero but it was wrong. what's the trick here
Notice that as you keep taking the value of the derivative at x = 0, you'll get 1, 1, 2, 6, etc. These are 0!, 1!, 2!, 3!,...(n-1)!, where n is the nth derivative you're taking. So the 10th derivative would be 9! = 362880. Hope that helps =)
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