Calc exam tomorrow...

**banshee.beat** · October 2nd 2008, 09:41 PM

Ok, exam tomorrow. I wanted to check some problems to see if I'm doing them right..

Implicit differentiation:
$tan(x-y)=\frac{y}{(1+x^2)}$
I got--
$y'=\frac{(1+x^2)^2sec^2(x-y)+2xy}{(1+x^2)+sec^2(x-y)}$

Is that correct?

Also, if you're supposed to prove the differentiability of a function at $a$ means that the function is continuous at $a$ , does that require just a $f(x)=f(a$ ) proof??

**Jameson** · October 2nd 2008, 10:01 PM

I'll just respond to the second part...

You're asked to prove that if F is a function from R->R, that $f'(a) \in R \rightarrow \lim_{x \rightarrow a}f(x)=f(a)$ .

If f is differentiable at a, by the limit definition of a derivative this means that $\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=f'(a)$

I don't know how detailed your proof is expected to be, but I would say that it follows from this limit definition of the derivative that $\lim_{x \rightarrow a} a \in R$ . To be continuous at a certain point, the limit must exist at a, and f(a) must be defined (meaning no holes or something like that). Thus again by the limit above, f(a) must exist otherwise the derivative would not.

**banshee.beat** · October 2nd 2008, 10:15 PM

Originally Posted by Jameson

I'll just respond to the second part...

You're asked to prove that if F is a function from R->R, that $f'(a) \in R \rightarrow \lim_{x \rightarrow a}f(x)=f(a)$ .

If f is differentiable at a, by the limit definition of a derivative this means that $\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=f'(a)$

I don't know how detailed your proof is expected to be, but I would say that it follows from this limit definition of the derivative that $\lim_{x \rightarrow a} a \in R$ . To be continuous at a certain point, the limit must exist at a, and f(a) must be defined (meaning no holes or something like that). Thus again by the limit above, f(a) must exist otherwise the derivative would not.

The proof is mainly getting the derivative of f(x) using the definition of a limit. I know how to do that.. I just wasn't quite sure if continuity was implied through that. Thank you!

Can anyone check the first part for me? That was the hardest one I could find in that chapter.. I feel like if I can do that one, I can do any of the implicit diff. on the exam.

**Jameson** · October 2nd 2008, 10:27 PM

Originally Posted by banshee.beat

The proof is mainly getting the derivative of f(x) using the definition of a limit. I know how to do that.. I just wasn't quite sure if continuity was implied through that. Thank you!

You probably know this, but just to be sure, make sure you don't mean in the above quote that if f is a function going from x to y that $f(a) \in y \rightarrow f'(a) \in y$ holds for any function. Only the opposite. I'll check the first part. I was hoping someone else would, but I'll break out a pencil and paper.

**Jameson** · October 2nd 2008, 10:58 PM

Ok, I'm getting something different, just in the final answer of the denominator. Since I'm prone to making algebra errors, I'll post my steps and you can check them/check them against your own if you believe them to be correct.

$\sec^2(x-y)(1-y')=\frac{(1+x^2)y'-2xy}{(1+x^2)^2}$

Now multiply through by the denominator of the RHS and multiply everything by -1 to make it nicer.

$\sec^2(x-y)(1+x^2)^2(y'-1)+(1+x^2)y'=2xy$

Multiply out the first term in the LHS and then factor out a y'. Multiple steps here...

$y' \left[\sec^2(x-y)(1+x^2)^2+(1+x^2) \right] = 2xy+\sec^2(x-y)(1+x^2)^2$

Finally divide by the denom. if LHS to solve for y' and I get:

$y' = \frac{\sec^2(x-y)(1+x^2)^2}{[1+x^2](\sec^2(x-y)(1+x^2)+1)}=\frac{\sec^2(x-y)(1+x^2)}{\sec^2(x-y)(1+x^2)+1}$

So look that over, and if you spot something that doesn't make sense I'll either correct it or explain it.

**banshee.beat** · October 2nd 2008, 11:41 PM

Okay. So I guess I don't really understand what happened to the 2xy, but other than that what you did makes enough sense. If you don't mind.. could you tell me what I did wrong with my algebra? Here are the steps I took.

I did it a little bit differently..
$tan(x-y)=\frac{y}{1+x^2}$

Diff.
$sec^2(x-y)(1-y')=\frac{(1+x^2)(y')-(y)(2x)}{(1+x^2)}$

Distribute, divide..
$(1+x^2)^2sec^2(x-y)-sec^2(x-y)(y')=(1+x^2)(y')-2yx$

Move stuff around..
$(1+x^2)^2sec^2(x-y)+2xy=(1+x^2)(y')+sec^2(x-y)(y')$

Factor and divide..
$y'=\frac{(1+x^2)^2sec^2(x-y)+2xy}{(1+x^2)+sec^2(x-y)}$

I'll probably check this sometime in the morning before the exam. I've got 8am chem lecture. =) I'll get a lovely 4hrs of sleep...

**Jameson** · October 3rd 2008, 12:12 AM

#1) Your wrote: $sec^2(x-y)(1-y')=\frac{(1+x^2)(y')-(y)(2x)}{(1+x^2)}$ .

The denominator should be squared. You treated it as such in later steps so I assume this is just a typo.

#2) $(1+x^2)^2sec^2(x-y)-sec^2(x-y)(y')=(1+x^2)(y')-2yx$

You made an error here.

When you multiply the denom. of the RHS you get...

$(1+x^2)^2\sec^2(x-y)(1-y')=(1+x^2)y'-2xy$

Now when distributing the LHS you get...

$(1+x^2)^2\sec^2(x-y)-(1+x^2)\sec^2(x-y)y'=...$

You've made a $(1+x^2)^2$ disappear.

All steps after are wrong based on this. Fix it and get back to me.

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