-
Members of the previous forum can retrieve their temporary password here, (login and check your PM).
calculus problem
- Thread starter angryanus
- Start date
[code:1:3395a653ce] f g * f' - f * g' --- => --------------- g g^2 x + 2 (3x - 4)^3 - (x + 2) * 3(3x - 4)^2 * 3 ---------- => -------------------------------------- (3x - 4)^3 (3x - 4)^6 3x - 4 - 9x - 18 -6x - 22 = ---------------- = ---------- (3x - 4)^4 (3x - 4)^4 [/code:1:3395a653ce] I hope it helps.
[quote:3acefa6c20="Doerak"][code:1:3acefa6c20] f g * f' - f * g' --- => --------------- g g^2 x + 2 (3x - 4)^3 - (x + 2) * 3(3x - 4)^2 * 3 ---------- => -------------------------------------- (3x - 4)^3 (3x - 4)^4)^2 3x - 4 - 9x - 18 -6x - 22 = ---------------- = ---------- (3x - 4)^4 (3x - 4)^4 [/code:1:3acefa6c20] I hope it helps.[/quote:3acefa6c20] OOoooo ok i didnt use the quotient rule before. I understand how to get the top part but now i get tripped up on how the power becomes ^4 for the bottom.
*Edit Doerak: Image deleted due to incorrectness (see below next reply for the correction)* Divide the upper factors by the lower factors. This explains the blue correction. I can't supply you with correct English maths terms as English is not my native language. I hope the maths does the talk. I'll make some extra steps to explain how the blue corrections work out: [code:1:4c3e8a4680] (3x - 4)^1 - (x + 2) * 3(3x - 4)^0 * 3 (3x - 4)^1 = 3x -4 -------------------------------------- : (3x - 4)^0 = 1 (3x - 4)^4 3x - 4 - (x + 2) * 3 * 1 * 3 3x - 4 - 9(x + 2) 3x - 4 - 9x + 18 = ---------------------------- = ----------------- = ----------------- (3x - 4)^4 (3x - 4)^4 (3x - 4)^4 [/code:1:4c3e8a4680]
[quote:800e3b3420="angryanus"]How does the ((3x-4)^'3')^2 become ((3x-4)^'4')^2? Note the '3' and '4'. I have thought about it and I still don't understand what I'm doing wrong. :? When I do it I think it becomes ((3x-4)^3)^2 = 3x-4^6 but then itis not the right answer.[/quote:800e3b3420] In my attempt to explain this as simply as possible I made a terrible, TERRIBLE mistake... [code:1:800e3b3420] (3x - 4)^3 - (x + 2) * 3(3x - 4)^2 * 3 (3x - 4)^3 - (x + 2) * 3(3x - 4)^2 * 3 -------------------------------------- = -------------------------------------- (3x - 4)^3)^2 (3x - 4)^6 [/code:1:800e3b3420] You have to write (3x - 4)^4)^2 all out => (3x - 4)^6. You can not just delete the ^2, because it's totally different. This is what happens to the exponent: 3 * 2 - 2 (= 4). This is not the same as 3 * (2 - 2). The rest of the solution is correct (so is the answer). Another way of solving the problem: [code:1:800e3b3420] f * g => f' * g + f * g' x + 2 ---------- = (x + 2)(3x - 4)^(-3) (3x - 4)^3 => (3x - 4)^(-3) + (x + 2) * -3 * (3x -4)^(-4) * 3 1 9(x + 2) (3x - 4) 9(x + 2) = ---------- - ---------- = ---------- - ---------- (3x - 4)^3 (3x - 4)^4 (3x - 4)^4 (3x - 4)^4 3x - 4 - 9x - 18 -6x - 22 = ---------------- = ---------- (3x - 4)^4 (3x - 4)^4 [/code:1:800e3b3420]
sorry to bother with math again, but this time instead of calculus it is stoichiometry. here is my problem... when i use my ph meter to test my base solution it gives me weird numbers. the ph jumps to 12ish and very slowly climbs up but never gets to 13. i believe that my ph meter needs calibrating or something. anyways i decided to try and calculate how many grams naoh would be needed in a a 480ml solution. i used ~43 grams and when i do the math, the ph comes to be ~14.4. can someone check my answer as i dont trust myself :lol: