Meaghan26 said:
Compute the 9th derivative of:
arctan((x^3)/2) at x=0
f^9(0)=?
As I describe this, keep the formula for a Maclaurin series handy. You wrote
arctan(x) =x-x^3/3 + x^5/5 +...
If you compare this with the formula for Maclaurin series, you'll see that this tells you that,
at x = 0,
first derivative of arctan(x) = 1 ,
second derivative = 0 ,
third derivative = -1/3 ,
fourth derivative = 0 ,
fifth derivative = +1/5 , etc.
What you have is f(x) = arctan[ (x^3)/2 ] = arctan [ u(x) ] , which is a composite function. So we use the Chain Rule to find
f'(x) = [ d (arctan u) / du ] · u' .
------- EDIT: PLEASE DISREGARD THIS - THE DERIVATION IS IN ERROR ----------
Okay, we have a way to go yet -- this is a product, so the next derivative is
f''(x) = [ d^2 (arctan u) / du^2 ] · u'
+ [ d (arctan u) / du ] · u'' .
[
EDIT: this should be f''(x) = [ d^2 (arctan u) / du^2 ] ·
(u')^2
+ [ d (arctan u) / du ] · u'' .
What follows will be more complicated than I believed at the time I posted this. My apologies.]
Now, at this point, most people would start screaming, "And we have to go to the
ninth derivative!?" . But we hit bottom with terms
because
u = (x^3)/2 , u' = 3(x^2)/2 , u'' = 3x , u''' = 3 , u^(4) and all higher derivatives = 0.
We also start to notice something about the derivatives of f(u) -- the next derivative is
f'''(x) = [ d^3 (arctan u) / du^3 ] · u'
+ 2 · [ d^2 (arctan u) / du^2 ] · u''
+ [ d (arctan u) / du ] · u'''
The terms start to look like the sort of thing we get from calculating the binomial power
(x+y)^n . The coefficients are the terms in Pascal's Triangle or the combinatorial values nCr and the total of the level of derivatives we take is one higher than the derivative of f(x) we are working on.
This suggests that the ninth derivative of f(x) looks like
f^(9) (x) = [ d^9 (arctan u) / du^9 ] · u'
+ 9 · [ d^8 (arctan u) / du^8 ] · u''
+ (9·8/2) · [ d^7 (arctan u) / du^7 ] · u''' ;
we don't need any more terms because the derivatives of u beyond that are all zero.
Now we can go back to the Maclaurin series. The seventh derivative at x = 0 is -1/7 , the eighth derivative is 0, and the ninth derivative is +1/9 . But u(0) = (0^3)/2 = 0 , so these will also be the values of the derivatives of arctan(u). This already reduces our expression to
f^(9) (0) = [ d^9 (arctan u) / du^9 ] · u'
+ 9 · [ d^8 (arctan u) / du^8 ] · 0
+ (9·8/2) · [ d^7 (arctan u) / du^7 ] · u'''
= (1/9)·u' + (36)·(-1/7)·u'''
At x = 0, u'(0) = 3(0^2)/2 = 0 , and we already know that u''' = 3 . Thus,
f^(9) (0) = (1/9)·0 + (36)·(-1/7)·3 = -108/7 .
After all this terribly rigorous agony, this suggests something we (to say, I) might have noticed from the start: at x = 0 , the
only derivative of u which
isn't zero is u''' = 3 . So we could just have looked at the pattern of derivatives for f^(n) (0) and, after writing the result for f^(9) = 0 , noticed that the only term that matters is
(9·8/2) · [ d^7 (arctan u) / du^7 ] · u''' ; absolutely everything else disappears.
[
EDIT: The basic idea of what happens at x = 0 is still correct, but the coefficient is not correctly found as described here.]
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I believe something similar happens for your other problem
(cos(6x^2)-1)/(x^2) at x=0
f^6(0)=?
I'm reading this as
[ cos(6·x^2) - 1 ] / (x^2)
Look at the Maclaurin series for cos x; divide it by x^2. The 1/(x^2) term in the series will get canceled out by the -1/(x^2) term in your function. The coefficients in your series now start with the second derivative of cos x , so you can read off what derivative values you need.
This time, u = 6·(x^2): how high do the derivatives go? Which aren't zero at x=0?
Again here, u(0) = 0, so the derivatives of cosine at u = 0 are the same as the derivatives at x = 0.
Use the method we described to find the non-zero terms for f^(6) (0) and calculate the result.
