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

- IsTim

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

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- katieb

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- IsTim

|dw:1335261368315:dw|

- IsTim

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

- IsTim

|dw:1335261446849:dw|

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## More answers

- chaise

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?

- IsTim

Deriatives is f'(x) right? we're not doing that yet.

- IsTim

Methods we're using is substitution, factoring and the conjugate method.

- IsTim

I'm using the Calculus Phobe videos from Calculus Help (my teacher showed me the website and told me to use that).

- chaise

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?

- chaise

dy/dx =lim (h-->0) (f(x+h)-f(x))/h
rather.

- IsTim

Yes. That's what I did.

- IsTim

But I can't go any further. |dw:1335262087563:dw| That's the furthest I got up to.

- chaise

Expand your brackets and factor out a h.

- IsTim

The bottom one to? I was told not too.

- chaise

Not sure - I'm not familiar with the long method, I just use the shortcut :)

- Callisto

Can you take derivatives? Must you use first principle to find it?

- IsTim

Derivatives is our next unit.

- IsTim

I can't isolate or factor out anything. I must had made a mistake beforehand I guess.

- IsTim

Anyone?

- Callisto

I hope you don't mind me posting my solution here

##### 1 Attachment

- IsTim

I guess not, but it's notm much like my work.

- Callisto

Oh shoot... I got some mistakes there! Wait

- Callisto

Once again :(

##### 1 Attachment

- Callisto

Oh my dear!!! The final answer should be -4/3 ... not -2 as I've written :S

- Callisto

Is your question \[\frac{3}{x^2-1}\] or \[\frac{3}{x^2+1}\]?

- IsTim

First one.

- IsTim

Maybe I should upload my work, and maybe you check for problems?

- Callisto

Sure!

- IsTim

IT's c.

##### 2 Attachments

- IsTim

Starting from c) at Scan001.

- Callisto

f(a) is not 1

- Callisto

f(a) is 3/ (a^2 -1)

- IsTim

but since x=2, I subbed in a=2 into that equation...

- Callisto

for the denominator, (a+h)-a =h , not a+h -2
You need to find the limit before substituting the numbers.

- Callisto

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

- IsTim

I thought the unknown to be m, and that h was still the variable withint the equation?

- Callisto

h is something you need to deal with.
Should I upload an example to should how you do first principle for you ?

- IsTim

I've got an example here with me from my teacher.

- Callisto

If you don't need it, it's okay :)
Should I keep checking for you or you want to do corrections first?

- IsTim

What should I correct?

- Callisto

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)

- IsTim

Sorry if I sound ignorant, but that's how I saw it in the example...the h is alone.

- Callisto

Do you mind uploading your example also?

- IsTim

##### 1 Attachment

- Callisto

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...

- IsTim

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.

- Callisto

It's good that you notice the difference :)

- IsTim

Just looking through my notes I guess. I've got a test coming up, so I thought reviewing would be best.

- Callisto

Yup :)

- Mimi_x3

well, where do you need help with?
This is pretty long..

- IsTim

Maybe I should just leave this question. I tried this for over 2 hours by now.

- Mimi_x3

First, what is the question?
\[\frac{1}{x-2} \]??
or
\[\frac{3}{x^{2}-1} \]??

- IsTim

Second one, whereas x=2.

- IsTim

Ok. Thanks for the attempt guys. I'll try other problems, and maybe this might make more sense.

- Mimi_x3

wait..i will try and do it first..
can't you take the derivative which is easier?

- IsTim

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.

- Mimi_x3

Hmm..I might try it but it might take a while since i havent done these in a while...

- IsTim

Ok. Well, I guess you can do so, if yo uhave free time. But most likely you too have something important coming up (immediately).

- Mimi_x3

No, I don't have anything important..i might be able to do it but not sure..lol

- Mimi_x3

I have to review these anyway

- apoorvk

^^That was mammoth. phew. Istim do you know how to differentiate using chain rule?

- Mimi_x3

lol, i failed...do you remember how to do these apoor? my brain is failling me :(

- IsTim

No, I don't (Sorry, cleaning up house).

- apoorvk

Hmm. so you want to do this by the first principle of finding derivatives okay.

- IsTim

I am just oh so selective in the method used. That be the end of me.

- Mimi_x3

\[\large\frac{f(x+h)-f(x)}{h} =>\frac{\frac{3}{(x+h)^{2}-1}-\frac{3}{x^{2}-1}}{h} \]

- Mimi_x3

right so far?

- IsTim

I have no idea.

- chaise

should be the limit as h approaches zero, but, apart from that, correct. ^_^

- Mimi_x3

\[\large =>\frac{\frac{3}{h^{2}+2hx+x^{2}-1}-\frac{3}{x^{2}-1}}{h} \]
\[\large >\frac{\frac{3(x^{2}-1)-3(h^{2}+2hx+x^{2}-1)}{(h^{2}+2hx+x^{2}-1)(x^{2}-1)}}{h} \]

- Mimi_x3

I think that im doing it wrong..i dont think that its this complicated..

- chaise

Perhaps you should just use quotient rule.

- Mimi_x3

but it says derivatiive thing is not allowed..

- IsTim

Before using deratives. Referencing Calculus Phobe from Calculus Help. I think I used Conjugate Method.

- Mimi_x3

whaat..im 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.

- Mimi_x3

\[\frac{3}{x^{2}-1} => 3(x^2-1)^{-1}\]
\[-3(x^2-1)^{-1}*2x \]
\[-6x(x^2-1)^{-1}\]
easier i suppose..

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