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Antiderivative of Log
tankia:
--- Quote from: "coblin" ---
--- Quote from: "tankia" ---Whyyy? Lol ....I'm so screwed for exams ><;
--- End quote ---
dy/dx = 1/x, find y: for R\{0}
Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.
Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.
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_.... You lost me at dy/dx >_>;
Odette:
--- Quote from: "tankia" ---
--- Quote from: "coblin" ---
--- Quote from: "tankia" ---Whyyy? Lol ....I'm so screwed for exams ><;
--- End quote ---
dy/dx = 1/x, find y: for R\{0}
Since 1/x is defined for all values except 0, the antiderivative (and the derivative) should also exist for those values.
Notice that ln(x) is only defined for x>0. To correct for this, we place: ln|x| so that it is defined for R\{0}. That's a floppy argument, and what I'm about to suggest is still floppy, but if you look at ln|x| for x is less than zero, you'll notice that it is decreasing. Look at 1/x, it is the derivative of ln|x| and for values less than zero, it has negative numbers! It makes sense.
--- End quote ---
_.... You lost me at dy/dx >_>;
--- End quote ---
Tanika I'll explain it to you on msn =]
tankia:
Okay :D
Freitag:
It seriously doesn't matter too much :p
tankia:
--- Quote from: "Freitag" ---It seriously doesn't matter too much :p
--- End quote ---
LOL. I was doing my FIRST practice exam (one that i actually finished...Thank the lord for solutions :D) and i finally got what you guys were talking about...kinda...some what..maybe...maybe not...Okay...Yea..I'll go back to procrastinating :D
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