y=e^xlnx
y=ln(3x^2)
y=lnx/x^2+1

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- anonymous

y=e^xlnx
y=ln(3x^2)
y=lnx/x^2+1

- schrodinger

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- anonymous

what's the question? which equation is nicest?

- anonymous

What's the goal? Solve for x in this system of equations?

- anonymous

heh, I like the first one ;)

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## More answers

- anonymous

me too! :)

- anonymous

each one is a different question its says compute the derivatives of each function

- anonymous

Oh, OK, are you familiar with (1.) product rule and (2.) quotient rule and the (3.) power rule?

- anonymous

yeah a little

- anonymous

(1.) \[y=e^x*ln(x)\]
(2.) \[y=ln(3x^2)\]
(3.) \[y=\frac{ln(x)}{x^2}+1\]

- anonymous

And you know chain rule? a little, right? ;)

- anonymous

the 1st equation is \[y=e ^{xlnx}\]

- anonymous

yes

- anonymous

OK, well, let's work with equation 2, for me that is the easiest one. We are differentiating with respect to y, right?

- anonymous

yes

- anonymous

Cool, \[y=ln(3*x^2)\] We will use the Chain Rule here

- anonymous

\[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\]
\[u = 3*x^2\]and \[\frac{d*ln(u)}{du} = \frac{1}{u}\]
\[\frac{du}{dx} = 3*2*x^{2-1} = 6x\]

- anonymous

Yes, and to get du/dx we use the power rule

- anonymous

\[x^y = y*x^{y-1}\] is the power rule

- anonymous

Can you finish solving the equation?

- anonymous

ok cause when he explained it in class i did not understand it at all.

- anonymous

All right, how is it now?

- anonymous

ok so for the power rule what do i plug in or do I have to plug anything in

- anonymous

Nah, you don't need to plug-in anything. So, what do you have so far? If you are writing it down, you can snap a picture with your phone if it has a camera, then post the picture here

- anonymous

\[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\]
\[\frac{d*ln(u)}{du} = \frac{1}{u}\]
\[\frac{du}{dx} = 3*2*x^{2-1} = 6x\]
\[u=3*x^2\]
So you have \[\frac{1}{3*x^2} * 6x\]
Simplify and you're done with that equation

- anonymous

O ok I didn't know whether what I had was right but thats what I just got to.

- anonymous

This maybe the hardest of the bunch because of the substitution, chain rule, and product rule

- anonymous

yea It was

- anonymous

Well, because substitution can be confusing because you introduce a new variable to differentiate on

- anonymous

*or with respect to

- anonymous

O ok

- anonymous

Now (1.)\[y=e^x * lnx\]

- Callisto

Excuse ... me.... are we finding dy/dx or dx/dy?

- anonymous

dy/dx I believe

- Callisto

Thank you. Please continue....

- anonymous

Heh, thanks, are you following along @Callisto ?

- anonymous

\[y=e^x* lnx\]
Here we use the product rule. We can think of this equation as the product of 2 functions of x

- anonymous

\[(f*g)' = f'*g + f*g'\]
So the derivative of the product of these functions, f and g
Is the sum of the derivative of the first * the second + the first * the derivative of the second

- Callisto

For the second one, I was thinking in this way..
\[y=ln(3x^2)=ln3+ lnx^2 = ln3+2lnx\]\[\frac{dy}{dx}=\frac{d}{dx}2lnx = ...\]

- anonymous

Yes, that is easier! @hooverst If you did it that way you don't have do substitution

- anonymous

Ok dpflan

- anonymous

It can be really fun in math the find the simpler way of doing things, so you just use some intuition and understanding to make your work easier, simpler can be much more beautiful too

- anonymous

*to find

- anonymous

hold on... i think @hooverst clarified that the first equation is:
\(\large y=e^{xlnx} \) , not \(\large y=e^{x}\cdot lnx \)

- anonymous

Heh, you're right, man

- anonymous

so callisto when you solved your equation for the second 1 what did you get for your anserw

- anonymous

Thanks, getting a little carried away. Let me step away for a bit

- anonymous

ur doing fine....:)

- Callisto

@hooverst What did you get ??

- anonymous

is the anserw \[y=\ln(3x^2)=1\div 3x^2\]

- Callisto

Not really...
what is \(\frac{d}{dx}lnx\) ?

- anonymous

im not sure

- anonymous

are we still working on the second equation? @hooverst ???

- anonymous

because you can do this many different ways... but in the end, the derivative should be the same...

- anonymous

yes but I was trying to figure out was 2lnx the anserw for #2 or is y=ln(3x^2)=1/3x^2 the anserw

- anonymous

answer as in the derivative? you and @dpflan , worked it out to \(\large \frac{6x}{3x^2} \) simplified to....???

- anonymous

in the case with @Callisto , her method was to simplify the function first:
\(\large y=ln(3x^2)=ln3+2lnx \)
so
\(\large y'=[ln3]'+[2lnx]'= \)
...???

- anonymous

and either way the derivative is the same...

- Callisto

The key point is you need to know what \(\frac{d}{dx}lnx\) is.

- anonymous

Yes, that is a derivative you need to memorize, it will be quite useful

- anonymous

Ok so many of you have said had your own opinion about the equation so which one is the right 1.

- anonymous

Hehe, right, so, what is your opinion? ;p
You can solve a problem many, many, different ways

- anonymous

For you, how would you approach it now that you've seen how we would?

- Callisto

I would say, none of us have given you the final answer. But we have given you some steps you need using different approaches. Though, the final answer will be the same.

- anonymous

Ok Think the one Callisto gave me is more simple for #2

- anonymous

Definitely, that was an awesome application of intuition

- anonymous

well, and mathematical understanding

- anonymous

i guess it comes down to preference because i personally would've used u'/u ... the method used earlier...

- Callisto

The step I left for you is to find \(\frac{d}{dx}lnx\). Multiply the answer you get for that by 2. Then, it's done.

- anonymous

ok

- Callisto

May I ask you again- what is d/dx ( lnx) ?

- anonymous

would you multiply it.

- Callisto

Nope....
Hint: look it up in the pdf.: http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_Derivatives.pdf
You can find the answer for d/dx (lnx) there.

- anonymous

Ok is it d/dx(ln(x))=1/x

- Callisto

Yes.
dy/dx = d/dx (ln3 + lnx^2) = d/dx (2lnx) = 2 d/dx(lnx) = ...?

- anonymous

is it 2/x or just 2x Idon't know if its right

- Callisto

Which one do you think? (i) 2/x (ii) 2x

- anonymous

2/x

- Callisto

Yes. That's correct. Any questions?

- anonymous

just 1 when we started simplifying the equation where did you get the 2 from?

- Callisto

\[lnx^a = a\ln x\]

- anonymous

Ok I get it. thanks Callisto

- Callisto

Welcome.

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