Compute the 10th derivative of f(x) = arctan (x^2/(6)) Hint: Use the MacLaurin s

Question

Compute the 10th derivative of

f(x) = arctan (x^2/(6))

Hint: Use the MacLaurin series for f(x)

Look. I know the answer is 280/3. I also understand that we take the MacLaurin (Taylor) expansion for f(x)=arctan (x) and then substitute x^2/(6) for the x value in the expansion for arctan (x). I am not understanding what computing the 10th derivative at x=0 means. I really need a detailed explanation as to what exactly I am doing in this problem and what is it I am looking for.

Please, I don't need detailed answer showing the expansion of the arctan.(x) series, nor the fact that we drop in x^2/6. I get that, for example, the 4th term of arctan (x^2/6) = ((-x^2/6)^5)/(5)...but now what? Do I extend it out to the tenth term and plug in 0 for x at that term? That makes no sense...So I am confused by what exactly the problem is asking me to do, and how do I do it, after I come up with the 10th term.

I really am looking for an explanation beyond what I know so far

Explanation / Answer

given: y = arctan(z)

you get the expansion of = z - (z^3)/3 + (z^5)/5 + ................ infinite terms

GIVEN f(x) = arctan (x^2/(6))

so MACLAURIN SERIS FOR GIVEN f(x) IS PUT Z=X^2/6 IN THE ABOVE MENTIONED EXPRESSION

f(x) = X^2/6 - ((x^2/6)^3)/3 + ((X^2/6)^5)/5 + ................ infinite terms

AS YOU CAN SEE THIS IS A POLYNOMIAL IN X

NOW WE HAVE TO CALCULATE THE 10TH DERIVATIVE

WE KNOW d/dx(X^N) = N *X^(N-1)

ALSO THE POLYNOMIAL f(x) IS A POLYNOMIAL IN POWERS OF X

SO IF YOU CALCULATE THE 10TH DERIVATIVE, ONLY THE COEFFICIENT OF THE X^10 WILL BE LEFT

ALL THE 10TH DERIVATIVES OF X^n FOR N<10 WILL GO TO ZERO

ALL THE 10TH DERIVATIVES OF X^n FOR N=10 EVALUATED AT X=0 WILL BE ALSO ZERO

SO x^10 TERM IN THE ABOVE EXPANSION = ((X^2/6)^5)/5 = X^10/ (5*6^5) = X^10/(38880)

IF U TAKE THE DERIVATIVE 10 TIMES OF THE ABOVE TERM YOU ARE LEFT WITH 10!/38880 = 280/3

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