Question

39th derivative of Sin(3x^2) ?

Answer 1

more specificly
f^39(0)

Answer 2

No, it's more complicated than that... haha

Answer 3

Nice question. You obviously want to try and find an expression for it without explicitly differentiating 39 times. So write down the first 5 or so derivatives and see.

f(x) = sin(3x^2) f(0) = 0
f' = 6x cos(3x^2), f'(0) = 0
f'' = 6 cos(3x^2) + 36x^2 sin(3x^2), f''(0) = 6
f''' = ... etc.

It should be obvious that any sin term is zero and any any cos term multiplied by a non-zero power of x is also zero. So see if you can figure out the recursive expression and then the general expression for \( f^{(n)}(0) \).

Answer 4

im doing the same thing

Answer 5

d^39/dx^39(sin(3 x^2)) = 1169352761683327101078941778119884800000 x sin(3 x^2)-2227915756473955677973140996096 x^39 cos(3 x^2)-275147595924533526229682913017856 x^37 sin(3 x^2)+15270691573811610705747401672491008 x^35 cos(3 x^2)+504781193689883798328872444174008320 x^33 sin(3 x^2)-11105186261177443563235193771828183040 x^31 cos(3 x^2)-172130387048250375230145503463336837120 x^29 sin(3 x^2)+1941248253933045898428863177947632107520 x^27 cos(3 x^2)+16223288979297597865441213701419496898560 x^25 sin(3 x^2)-101395556120609986659007585633871855616000 x^23 cos(3 x^2)-475056957379894937494979984543881101312000 x^21 sin(3 x^2)+1662699350829632281232429945903583854592000 x^19 cos(3 x^2)+4307902863513138183193113950750194532352000 x^17 sin(3 x^2)-8137149853302594346031437462528145227776000 x^15 cos(3 x^2)-10953855571753492388888473507249426268160000 x^13 sin(3 x^2)+10171437316628242932539296828160181534720000 x^11 cos(3 x^2)+6215878360161704014329570283875666493440000 x^9 sin(3 x^2)-2330954385060639005373588856453374935040000 x^7 cos(3 x^2)-479902373394837442282797705740400721920000 x^5 sin(3 x^2)+44435404943966429840999787568555622400000 x^3 cos(3 x^2)

Answer 6

...and this is why writing down the 39th derivative is not helpful

Answer 7

there is an x term in each of the terms listed above, therefore, th thiry-ninth derivative at 0 equals 0.

Answer 8

That's true, but the case could be made that one would not likely have the resources to find out that way unless they were sitting at a computer.

Answer 9

expand it taylor series, you will see that x only have even power(by 4)

sin(x)= x- x^3/3! + x^5/5!+.........

Sin(3x^2)= 3x^2 - 3^3 x^6/3! + 3^5 x^10 /5!

according taylor series
\[\sum _{n=0}^{\infty } f^n(0)\frac{x^n}{n!}\]

\[f^{39}(0)\frac{x^{39}}{39!}\]

but there is no x^39 term in our taylor series

Answer 10

looking at n=2\[f''(x)=xf'+6^2x^2f\]perhaps\[f^{(n)}(x)=xf^{n-1}+6^2x^2f^{n-2}\]

Answer 11

that would show that \[f^n(0)=0\]right?

Answer 12

Yes, your recessive formula isn't quite right.

Answer 13

aww :(

Answer 14

I liked Imran's proof.

Answer 15

oh yeah, took a moment for me to recognize his idea...
very cool

Answer 16

if you use this relation:\[\begin{align}
f&=e^{3ix^2}&=\cos(3x^2)+i\sin(3x^2)\\
f^1&=6ixe^{3ix^2}&=6ixf\\
f^2&=6i(xf^1+f)\\
f^3&=6i(xf^2+f^1+f^1)&=6i(xf^2+2f^1)\\
f^4&=6i(xf^3+f^2+2f^2)&=6i(xf^3+3f^2)\\
...\\
f^n&=6i(xf^{n-1}+(n-1)f^{n-2})
\end{align}\]so at x=0, we get:\[
f^n(0)=6i(0*f^{n-1}(0)+(n-1)f^{n-2}(0))=6i(n-1)f^{n-2}(0)
\]which cascades down to zero.

Answer 17

we only need to consider the imaginary terms in this.