Solve this problem to get a cookie:
Name:
Anonymous
2006-10-13 21:01
if f(x) = [-cos(sin(3x^2))]/[tan(3x)]
find f''(x)
(Yes this problem is quite tedious, but if you want the cookie, start working)
Name:
Anonymous
2006-10-13 21:28
f' = (sin(sin(3x^2))*cos(3x^2)*6x)/(tan(3x)) + (cos(sin(3x^2)))/(sin(3x)^2)
f'' = (cos(sin(3x^2))*cos(3x^2)^2*36x^2 - sin(sin(3x^2))*sin(3x^2)*36x^2 + sin(sin(3x^2))*cos(3x^2)*6)/(tan(3x)) + (sin(sin(3x^2))*cos(3x^2)*6x)/(sin(3x)^2) - (sin(sin(3x^2))*cos(3x^2)*6x)/(sin(3x)^2) - (cos(sin(3x^2))*cos(3x)*12)/(sin(3x))
I think.
Name:
Anonymous
2006-10-13 21:28
Also, just check it in Mathematica or something?
Name:
Anonymous
2006-10-13 22:43
>>1
I have a plate full of cookies in the kitchen already, sucker
Name:
Anonymous
2006-10-13 23:15
Name:
Anonymous
2006-10-14 6:52
>>2
Did you actually bother working that out, or did you use a program to do it?
You must really want that cookie...
Name:
Anonymous
2006-10-14 11:16
mathematica. does anyone know how to copy it without the backslashes? i had to take it out myself.
3 Cos[Sin[3 x^2]] Csc[3 x]^2 + 6 x Cos[3 x^2] Cot[3 x] Sin[Sin[3 x^2]]
Name:
Anonymous
2006-10-14 11:19
Name:
Anonymous
2006-10-14 11:22
f'' = (6 ((6 x^2 Cos[3 x^2]^2 Cos[Sin[3 x^2]] Cot[3 x] + Cos[3 x^2] ((Cot[3 x] - 6 x Csc[3 x]^2)) Sin[Sin[3 x^2]] - 3 Cot[3 x] ((Cos[Sin[3 x^2]] Csc[3 x]^2 + 2 x^2 Sin[3 x^2] Sin[Sin[3 x^2]])))))
Name:
Anonymous
2006-10-14 11:33
shorter lol
f'' = (18 Cos[Sin[3 x^2]] Cot[3 x] ((x^2 ((1 + Cos[6 x^2])) - Csc[3 x]^2)) + 6 ((Cos[3 x^2] ((Cot[3 x] - 6 x Csc[3 x]^2)) - 6 x^2 Cot[3 x] Sin[3 x^2])) Sin[Sin[3 x^2]])
Name:
Anonymous
2006-10-14 12:08
That's easily reduced to f'' = 1/0
Name:
Anonymous
2006-10-14 12:24
>>6
I worked it out by hand.
Name:
Anonymous
2006-10-14 22:33
maple outputs the following:
simplify(diff(f,x)):
3*(2*sin(sin(3*x^2))*cos(3*x^2)*x*sin(3*x)*cos(3*x)+cos(sin(3*x^2)))/sin(3*x)^2
Name:
Anonymous
2006-10-14 22:36
whoops, my bad, that was the first derivative,
simplify(diff(f,x$2));
-6*(-6*cos(sin(3*x^2))*cos(3*x^2)^2*x^2*cos(3*x)+6*cos(sin(3*x^2))*cos(3*x^2)^2*x^2*cos(3*x)^3+6*sin(sin(3*x^2))*sin(3*x^2)*x^2*cos(3*x)-6*sin(sin(3*x^2))*sin(3*x^2)*x^2*cos(3*x)^3-sin(sin(3*x^2))*cos(3*x^2)*cos(3*x)+sin(sin(3*x^2))*cos(3*x^2)*cos(3*x)^3+6*sin(sin(3*x^2))*cos(3*x^2)*x*sin(3*x)+3*cos(sin(3*x^2))*cos(3*x))/sin(3*x)^3
Name:
Anonymous
2006-10-16 18:41
God dammit, I'm doing this stuff in Maths now.
I'm 16 years old... Something challenging, please?
Name:
Anonymous
2006-10-16 20:38
>>15
Yes, calculate 1 / 0
Name:
Anonymous
2006-10-16 20:41
Name:
Anonymous
2006-10-17 1:24
Name:
Anonymous
2006-10-17 7:29
>>1
i found the 4 "x's" do i win?
Name:
Anonymous
2006-10-17 16:23
3Cos[Sin[3x^2]]Csc[3x]^2+6xCos[3x^2]Cot[3x]Sin[Sin[3x^2]]
Name:
Anonymous
2006-10-20 1:16
3·COS(SIN(3·x^2))/SIN(3·x)^2 + 6·x·COT(3·x)·COS(3·x^2)·SIN(SIN(3·x^2))
lol
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