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dpaInc Group TitleBest ResponseYou've already chosen the best response.0
what's the question? which equation is nicest?
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
What's the goal? Solve for x in this system of equations?
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
heh, I like the first one ;)
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
each one is a different question its says compute the derivatives of each function
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Oh, OK, are you familiar with (1.) product rule and (2.) quotient rule and the (3.) power rule?
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
yeah a little
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
(1.) \[y=e^x*ln(x)\] (2.) \[y=ln(3x^2)\] (3.) \[y=\frac{ln(x)}{x^2}+1\]
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
And you know chain rule? a little, right? ;)
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
the 1st equation is \[y=e ^{xlnx}\]
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
OK, well, let's work with equation 2, for me that is the easiest one. We are differentiating with respect to y, right?
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Cool, \[y=ln(3*x^2)\] We will use the Chain Rule here
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\] \[u = 3*x^2\]and \[\frac{d*ln(u)}{du} = \frac{1}{u}\] \[\frac{du}{dx} = 3*2*x^{21} = 6x\]
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Yes, and to get du/dx we use the power rule
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
\[x^y = y*x^{y1}\] is the power rule
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Can you finish solving the equation?
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
ok cause when he explained it in class i did not understand it at all.
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
All right, how is it now?
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
ok so for the power rule what do i plug in or do I have to plug anything in
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Nah, you don't need to plugin 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
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\] \[\frac{d*ln(u)}{du} = \frac{1}{u}\] \[\frac{du}{dx} = 3*2*x^{21} = 6x\] \[u=3*x^2\] So you have \[\frac{1}{3*x^2} * 6x\] Simplify and you're done with that equation
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
O ok I didn't know whether what I had was right but thats what I just got to.
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
This maybe the hardest of the bunch because of the substitution, chain rule, and product rule
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
yea It was
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Well, because substitution can be confusing because you introduce a new variable to differentiate on
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
*or with respect to
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Now (1.)\[y=e^x * lnx\]
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
Excuse ... me.... are we finding dy/dx or dx/dy?
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
dy/dx I believe
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
Thank you. Please continue....
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Heh, thanks, are you following along @Callisto ?
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
\[y=e^x* lnx\] Here we use the product rule. We can think of this equation as the product of 2 functions of x
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
\[(f*g)' = f'*g + f*g'\] 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
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
For the second one, I was thinking in this way.. \[y=ln(3x^2)=ln3+ lnx^2 = ln3+2lnx\]\[\frac{dy}{dx}=\frac{d}{dx}2lnx = ...\]
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Yes, that is easier! @hooverst If you did it that way you don't have do substitution
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Nice one @Callisto @hooverst So, you think you can solve equation (1.) with the product rule?
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
hold on... i think @hooverst clarified that the first equation is: \(\large y=e^{xlnx} \) , not \(\large y=e^{x}\cdot lnx \)
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Heh, you're right, man
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
so callisto when you solved your equation for the second 1 what did you get for your anserw
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Thanks, getting a little carried away. Let me step away for a bit
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
ur doing fine....:)
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
@hooverst What did you get ??
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
is the anserw \[y=\ln(3x^2)=1\div 3x^2\]
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
Not really... what is \(\frac{d}{dx}lnx\) ?
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
im not sure
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
are we still working on the second equation? @hooverst ???
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
because you can do this many different ways... but in the end, the derivative should be the same...
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
answer as in the derivative? you and @dpflan , worked it out to \(\large \frac{6x}{3x^2} \) simplified to....???
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
in the case with @Callisto , her method was to simplify the function first: \(\large y=ln(3x^2)=ln3+2lnx \) so \(\large y'=[ln3]'+[2lnx]'= \) ...???
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
and either way the derivative is the same...
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
The key point is you need to know what \(\frac{d}{dx}lnx\) is.
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Yes, that is a derivative you need to memorize, it will be quite useful
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
Ok so many of you have said had your own opinion about the equation so which one is the right 1.
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Hehe, right, so, what is your opinion? ;p You can solve a problem many, many, different ways
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
For you, how would you approach it now that you've seen how we would?
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
Ok Think the one Callisto gave me is more simple for #2
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
Definitely, that was an awesome application of intuition
 2 years ago

dpflan Group TitleBest ResponseYou've already chosen the best response.1
well, and mathematical understanding
 2 years ago

dpaInc Group TitleBest ResponseYou've already chosen the best response.0
i guess it comes down to preference because i personally would've used u'/u ... the method used earlier...
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
May I ask you again what is d/dx ( lnx) ?
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
would you multiply it.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
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.
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
Ok is it d/dx(ln(x))=1/x
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
Yes. dy/dx = d/dx (ln3 + lnx^2) = d/dx (2lnx) = 2 d/dx(lnx) = ...?
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
is it 2/x or just 2x Idon't know if its right
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
Which one do you think? (i) 2/x (ii) 2x
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
Yes. That's correct. Any questions?
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
just 1 when we started simplifying the equation where did you get the 2 from?
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.2
\[lnx^a = a\ln x\]
 2 years ago

hooverst Group TitleBest ResponseYou've already chosen the best response.0
Ok I get it. thanks Callisto
 2 years ago
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