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tan-¹√(x²-1)+csc-¹x, x>1
d/dx(tan-¹√(x²-1)+d/dx(csc-¹x)
This is where im confused:
I say:
For the first half:
(1/(1+x²-1))(0.5(x²-1)(2x))
(1/x²)(x³-x)
(x³-x)/x²
(x²-1)/x
The second half:
d/dx(csc-¹x)=d/dx(π/2-sec-¹x)
-1/(|x|√(x²-1))
Altogether:
((x²-1)/x)-(1/(|x|√(x²-1))
((x²-1)/x)-(1/(x√(x²-1))
My teacher says instead of (1/(1+x²-1))(0.5(x²-1)(2x)), you get (1/(1+x²-1))2x
One more question:
csc-¹x is just sinx right?
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I must be misreading the problem. It looks to me as though you are writing (for the first part):
[align=center]
[/align]In this case, by applying the chain rule I get:
[align=center]
[/align]or
[align=center]
[/align][align=center]
[/align]This is different from what both you and your teacher have, so perhaps I'm missing something. Anyway, csc is 1 over sin, not the inverse sine function, so you have to be careful (just like you can't write tan-1 as cos/sin and evaluate from there in the term on the left you get the wrong answer either way). Off the top of my head I can't remember a good technique for evaluating csc-1, so I'll take the wimpy way out and refer you to a tabulated answer
http://www.efunda.com/math/derivatives/TrigD.cfm
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One more question:
csc-¹x is just sinx right?
Not so fast, it might mean inverse the operation, like try to get the angle from the value, sort of like the other inverse
trigonometric functions. I don't know the history of the -1 symbol in this use. So I don't know, it may depend on the author.
It's either a backwards operation, or otherwise it is a reciprocal like you are suggesting, but I don't know the intention.
igloo myrtilles fourmis
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Yeah, that's definitely something you should be sure you're clear on. I would bet $20 your teacher means the "inverse of csc", not 1/csc, but you should definitely ask.
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k i did it again and i got wat u got i forgot to subtract 1 from the exponent(Stupid!), so the teacher is wrong?
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