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

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what's the question? which equation is nicest?
What's the goal? Solve for x in this system of equations?
heh, I like the first one ;)

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Other answers:

me too! :)
each one is a different question its says compute the derivatives of each function
Oh, OK, are you familiar with (1.) product rule and (2.) quotient rule and the (3.) power rule?
yeah a little
(1.) \[y=e^x*ln(x)\] (2.) \[y=ln(3x^2)\] (3.) \[y=\frac{ln(x)}{x^2}+1\]
And you know chain rule? a little, right? ;)
the 1st equation is \[y=e ^{xlnx}\]
OK, well, let's work with equation 2, for me that is the easiest one. We are differentiating with respect to y, right?
Cool, \[y=ln(3*x^2)\] We will use the Chain Rule here
\[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\] \[u = 3*x^2\]and \[\frac{d*ln(u)}{du} = \frac{1}{u}\] \[\frac{du}{dx} = 3*2*x^{2-1} = 6x\]
Yes, and to get du/dx we use the power rule
\[x^y = y*x^{y-1}\] is the power rule
Can you finish solving the equation?
ok cause when he explained it in class i did not understand it at all.
All right, how is it now?
ok so for the power rule what do i plug in or do I have to plug anything in
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
\[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\] \[\frac{d*ln(u)}{du} = \frac{1}{u}\] \[\frac{du}{dx} = 3*2*x^{2-1} = 6x\] \[u=3*x^2\] So you have \[\frac{1}{3*x^2} * 6x\] Simplify and you're done with that equation
O ok I didn't know whether what I had was right but thats what I just got to.
This maybe the hardest of the bunch because of the substitution, chain rule, and product rule
yea It was
Well, because substitution can be confusing because you introduce a new variable to differentiate on
*or with respect to
O ok
Now (1.)\[y=e^x * lnx\]
Excuse ... me.... are we finding dy/dx or dx/dy?
dy/dx I believe
Thank you. Please continue....
Heh, thanks, are you following along @Callisto ?
\[y=e^x* lnx\] Here we use the product rule. We can think of this equation as the product of 2 functions of x
\[(f*g)' = f'*g + f*g'\] 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
For the second one, I was thinking in this way.. \[y=ln(3x^2)=ln3+ lnx^2 = ln3+2lnx\]\[\frac{dy}{dx}=\frac{d}{dx}2lnx = ...\]
Yes, that is easier! @hooverst If you did it that way you don't have do substitution
Ok dpflan
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
*to find
Nice one @Callisto @hooverst So, you think you can solve equation (1.) with the product rule?
hold on... i think @hooverst clarified that the first equation is: \(\large y=e^{xlnx} \) , not \(\large y=e^{x}\cdot lnx \)
Heh, you're right, man
so callisto when you solved your equation for the second 1 what did you get for your anserw
Thanks, getting a little carried away. Let me step away for a bit
ur doing fine....:)
@hooverst What did you get ??
is the anserw \[y=\ln(3x^2)=1\div 3x^2\]
Not really... what is \(\frac{d}{dx}lnx\) ?
im not sure
are we still working on the second equation? @hooverst ???
because you can do this many different ways... but in the end, the derivative should be the same...
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
answer as in the derivative? you and @dpflan , worked it out to \(\large \frac{6x}{3x^2} \) simplified to....???
in the case with @Callisto , her method was to simplify the function first: \(\large y=ln(3x^2)=ln3+2lnx \) so \(\large y'=[ln3]'+[2lnx]'= \) ...???
and either way the derivative is the same...
The key point is you need to know what \(\frac{d}{dx}lnx\) is.
Yes, that is a derivative you need to memorize, it will be quite useful
Ok so many of you have said had your own opinion about the equation so which one is the right 1.
Hehe, right, so, what is your opinion? ;p You can solve a problem many, many, different ways
For you, how would you approach it now that you've seen how we would?
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.
Ok Think the one Callisto gave me is more simple for #2
Definitely, that was an awesome application of intuition
well, and mathematical understanding
i guess it comes down to preference because i personally would've used u'/u ... the method used earlier...
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.
May I ask you again- what is d/dx ( lnx) ?
would you multiply it.
Nope.... Hint: look it up in the pdf.: You can find the answer for d/dx (lnx) there.
Ok is it d/dx(ln(x))=1/x
Yes. dy/dx = d/dx (ln3 + lnx^2) = d/dx (2lnx) = 2 d/dx(lnx) = ...?
is it 2/x or just 2x Idon't know if its right
Which one do you think? (i) 2/x (ii) 2x
Yes. That's correct. Any questions?
just 1 when we started simplifying the equation where did you get the 2 from?
\[lnx^a = a\ln x\]
Ok I get it. thanks Callisto

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