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Can some of you calculus geniuses explain these problems a bit... maybe you'll find them fun or good for review. Oh yeah and Go Red Sox!!!

find derivatives for each:

sin7x
cos18x
sin9(cosx^4)
(x^2+5x-7)^8


find antiderivatives for each:

sin9x
(3x-5)^9


TROT NIXON HOME RUN!!!... sorry for randomness,heh, I'm a little excited.

By Frenchfries (Frenchfries) on Thursday, October 16, 2003 - 08:58 pm: Edit

I tried...sorry if they're wrong

7cos(7x)

-18sin(18x)

cos(36(x^3)(sin(x^40)))

(16x+40)(x^2+5x-7)^7

I don't know what antiderivatives are, we never did them in my class. Are you supposed to find f(x) if f'(x)=sin9x?

By Jason817 (Jason817) on Thursday, October 16, 2003 - 09:15 pm: Edit

"sin9x"

Try getting the integral in the form of
f(g(x))*g'(x)
so in this case, f is sin, g(x) is 9x, and we need the g'(x) to be the derivative of 9x, 9. This way, when you take the integral, g'(x) goes away.

Add in the 9 and multiple the outside by 1/9 to make up for it.

1/9 * integral(sin9x * 9 dx)=
1/9 * -cos(9x) + C
= -(1/9)cos(9x)+C

"(3x-5)^9 "

Same rule as above.
f is the ^9, g(x) is the 3x-5 and g'(x) has to be 3.

Add the 3 and put a 1/3 on the outside:
1/3 * integral((3x-5)^9 * 3 dx)=
(1/3)(1/10)(3x-5)^10 + C =
((3x-5)^10)/30) + C

By Bart_Simpson22 (Bart_Simpson22) on Thursday, October 16, 2003 - 09:54 pm: Edit

Thanks guys, still a bit iffy on the first four though, can anyone confirm french fries' work?

By Tanman (Tanman) on Thursday, October 16, 2003 - 10:15 pm: Edit

I got the same answers for three of the first 4 problems. On the one that is sin9(cosx^4) , if you mean sin(9cos(x^4)), then I'm getting -36(x^3)(sin(x^4))(cos(9(cos(x^4)))) by using chain rule 2 times. I might be wrong though..

By Abz1986 (Abz1986) on Thursday, October 16, 2003 - 11:28 pm: Edit

yea I'm getting the same for #3 as Tanman

By Volleygenius (Volleygenius) on Friday, October 17, 2003 - 01:14 pm: Edit

just a warning to all u guys. never drink and derive. hahahaha