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Find the equation of the tangent to the curve at the given x value.

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I got up to this:|dw:1335261401688:dw|

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Other answers:

Calculate the derivative of the equation, make the equation to equal, solve it and plug in your x value. Are you familiar with the quotient rule in derivatives?
Deriatives is f'(x) right? we're not doing that yet.
Methods we're using is substitution, factoring and the conjugate method.
I'm using the Calculus Phobe videos from Calculus Help (my teacher showed me the website and told me to use that).
Yes, that's correct. So you're differentiating using the definition? If you take the derivative of the equation y=3/(x^2-1) you can find the slope. if: f(x)=3/(x^2-1) then f(x+h)=3/((x+h)^2-1) dy/dx = (f(x+h)-f(x))/h Can you do the rest?
dy/dx =lim (h-->0) (f(x+h)-f(x))/h rather.
Yes. That's what I did.
But I can't go any further. |dw:1335262087563:dw| That's the furthest I got up to.
Expand your brackets and factor out a h.
The bottom one to? I was told not too.
Not sure - I'm not familiar with the long method, I just use the shortcut :)
Can you take derivatives? Must you use first principle to find it?
Derivatives is our next unit.
I can't isolate or factor out anything. I must had made a mistake beforehand I guess.
I hope you don't mind me posting my solution here
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I guess not, but it's notm much like my work.
Oh shoot... I got some mistakes there! Wait
Once again :(
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Oh my dear!!! The final answer should be -4/3 ... not -2 as I've written :S
Is your question \[\frac{3}{x^2-1}\] or \[\frac{3}{x^2+1}\]?
First one.
Maybe I should upload my work, and maybe you check for problems?
IT's c.
Starting from c) at Scan001.
f(a) is not 1
f(a) is 3/ (a^2 -1)
but since x=2, I subbed in a=2 into that equation...
for the denominator, (a+h)-a =h , not a+h -2 You need to find the limit before substituting the numbers.
After you've found the limit, you'll see the unknown x. By then, sub x=2 and you'll get the slope. But NOT when you're taking the limit
I thought the unknown to be m, and that h was still the variable withint the equation?
h is something you need to deal with. Should I upload an example to should how you do first principle for you ?
I've got an example here with me from my teacher.
If you don't need it, it's okay :) Should I keep checking for you or you want to do corrections first?
What should I correct?
Correct starting from the third line For numerator: a -> 3/ (a^2 -1) For denominator: a+h-2 -> h (in your previous step, a+h -a, simplify it , you'll get h)
Sorry if I sound ignorant, but that's how I saw it in the example...the h is alone.
Do you mind uploading your example also?
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First, the situation is different . You're given 2 values there, but only 1 here. Second, to be frank, I don't understand what you (or your teacher) is doing. It's not the normal way as I've learnt.. Third, I still insist that you shouldn't sub the value immediately...
Ok. I think I know what you are doing, and what my teacher is doing. You're going from a general formula, while my teacher is using the specific formula. The general formula doesn't sub in the x or a value until the end.
It's good that you notice the difference :)
Just looking through my notes I guess. I've got a test coming up, so I thought reviewing would be best.
Yup :)
well, where do you need help with? This is pretty long..
Maybe I should just leave this question. I tried this for over 2 hours by now.
First, what is the question? \[\frac{1}{x-2} \]?? or \[\frac{3}{x^{2}-1} \]??
Second one, whereas x=2.
Ok. Thanks for the attempt guys. I'll try other problems, and maybe this might make more sense.
wait..i will try and do it first.. can't you take the derivative which is easier?
Others suggested that, but that's for the next unit. I understand a bit on how that works, but I'm not allowed using it here.
Hmm..I might try it but it might take a while since i havent done these in a while...
Ok. Well, I guess you can do so, if yo uhave free time. But most likely you too have something important coming up (immediately).
No, I don't have anything important..i might be able to do it but not
I have to review these anyway
^^That was mammoth. phew. Istim do you know how to differentiate using chain rule?
lol, i you remember how to do these apoor? my brain is failling me :(
No, I don't (Sorry, cleaning up house).
Hmm. so you want to do this by the first principle of finding derivatives okay.
I am just oh so selective in the method used. That be the end of me.
\[\large\frac{f(x+h)-f(x)}{h} =>\frac{\frac{3}{(x+h)^{2}-1}-\frac{3}{x^{2}-1}}{h} \]
right so far?
I have no idea.
should be the limit as h approaches zero, but, apart from that, correct. ^_^
\[\large =>\frac{\frac{3}{h^{2}+2hx+x^{2}-1}-\frac{3}{x^{2}-1}}{h} \] \[\large >\frac{\frac{3(x^{2}-1)-3(h^{2}+2hx+x^{2}-1)}{(h^{2}+2hx+x^{2}-1)(x^{2}-1)}}{h} \]
I think that im doing it wrong..i dont think that its this complicated..
Perhaps you should just use quotient rule.
but it says derivatiive thing is not allowed..
Before using deratives. Referencing Calculus Phobe from Calculus Help. I think I used Conjugate Method. lost..i looked at your example but it got confusing..then i looked at my book its more confusing..using the chain rule is easier.
\[\frac{3}{x^{2}-1} => 3(x^2-1)^{-1}\] \[-3(x^2-1)^{-1}*2x \] \[-6x(x^2-1)^{-1}\] easier i suppose..

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