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dpaIncBest ResponseYou've already chosen the best response.0
what's the question? which equation is nicest?
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
What's the goal? Solve for x in this system of equations?
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
heh, I like the first one ;)
 one year ago

hooverstBest ResponseYou've already chosen the best response.0
each one is a different question its says compute the derivatives of each function
 one year ago

dpflanBest 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?
 one year ago

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

dpflanBest ResponseYou've already chosen the best response.1
And you know chain rule? a little, right? ;)
 one year ago

hooverstBest ResponseYou've already chosen the best response.0
the 1st equation is \[y=e ^{xlnx}\]
 one year ago

dpflanBest 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?
 one year ago

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

dpflanBest 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 \[\frac{d*ln(u)}{du} = \frac{1}{u}\] \[\frac{du}{dx} = 3*2*x^{21} = 6x\]
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
Yes, and to get du/dx we use the power rule
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
\[x^y = y*x^{y1}\] is the power rule
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
Can you finish solving the equation?
 one year ago

hooverstBest ResponseYou've already chosen the best response.0
ok cause when he explained it in class i did not understand it at all.
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
All right, how is it now?
 one year ago

hooverstBest 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
 one year ago

dpflanBest 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
 one year ago

dpflanBest 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
 one year ago

hooverstBest 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.
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
This maybe the hardest of the bunch because of the substitution, chain rule, and product rule
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
Well, because substitution can be confusing because you introduce a new variable to differentiate on
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
Now (1.)\[y=e^x * lnx\]
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Excuse ... me.... are we finding dy/dx or dx/dy?
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Thank you. Please continue....
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
Heh, thanks, are you following along @Callisto ?
 one year ago

dpflanBest 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
 one year ago

dpflanBest 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
 one year ago

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

dpflanBest 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
 one year ago

dpflanBest 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
 one year ago

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

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

hooverstBest 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
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
Thanks, getting a little carried away. Let me step away for a bit
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
@hooverst What did you get ??
 one year ago

hooverstBest ResponseYou've already chosen the best response.0
is the anserw \[y=\ln(3x^2)=1\div 3x^2\]
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Not really... what is \(\frac{d}{dx}lnx\) ?
 one year ago

dpaIncBest ResponseYou've already chosen the best response.0
are we still working on the second equation? @hooverst ???
 one year ago

dpaIncBest 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...
 one year ago

hooverstBest 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
 one year ago

dpaIncBest 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....???
 one year ago

dpaIncBest 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 \(\large y'=[ln3]'+[2lnx]'= \) ...???
 one year ago

dpaIncBest ResponseYou've already chosen the best response.0
and either way the derivative is the same...
 one year ago

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

dpflanBest ResponseYou've already chosen the best response.1
Yes, that is a derivative you need to memorize, it will be quite useful
 one year ago

hooverstBest 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.
 one year ago

dpflanBest 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
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
For you, how would you approach it now that you've seen how we would?
 one year ago

CallistoBest 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.
 one year ago

hooverstBest ResponseYou've already chosen the best response.0
Ok Think the one Callisto gave me is more simple for #2
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
Definitely, that was an awesome application of intuition
 one year ago

dpflanBest ResponseYou've already chosen the best response.1
well, and mathematical understanding
 one year ago

dpaIncBest 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...
 one year ago

CallistoBest 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.
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
May I ask you again what is d/dx ( lnx) ?
 one year ago

hooverstBest ResponseYou've already chosen the best response.0
would you multiply it.
 one year ago

CallistoBest 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.
 one year ago

hooverstBest ResponseYou've already chosen the best response.0
Ok is it d/dx(ln(x))=1/x
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Yes. dy/dx = d/dx (ln3 + lnx^2) = d/dx (2lnx) = 2 d/dx(lnx) = ...?
 one year ago

hooverstBest ResponseYou've already chosen the best response.0
is it 2/x or just 2x Idon't know if its right
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Which one do you think? (i) 2/x (ii) 2x
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Yes. That's correct. Any questions?
 one year ago

hooverstBest ResponseYou've already chosen the best response.0
just 1 when we started simplifying the equation where did you get the 2 from?
 one year ago

hooverstBest ResponseYou've already chosen the best response.0
Ok I get it. thanks Callisto
 one year ago
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