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anonymous
 4 years ago
y=e^xlnx
y=ln(3x^2)
y=lnx/x^2+1
anonymous
 4 years ago
y=e^xlnx y=ln(3x^2) y=lnx/x^2+1

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what's the question? which equation is nicest?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What's the goal? Solve for x in this system of equations?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0heh, I like the first one ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0each one is a different question its says compute the derivatives of each function

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, OK, are you familiar with (1.) product rule and (2.) quotient rule and the (3.) power rule?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(1.) \[y=e^x*ln(x)\] (2.) \[y=ln(3x^2)\] (3.) \[y=\frac{ln(x)}{x^2}+1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And you know chain rule? a little, right? ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the 1st equation is \[y=e ^{xlnx}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OK, well, let's work with equation 2, for me that is the easiest one. We are differentiating with respect to y, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Cool, \[y=ln(3*x^2)\] We will use the Chain Rule here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\] \[u = 3*x^2\]and \[\frac{d*ln(u)}{du} = \frac{1}{u}\] \[\frac{du}{dx} = 3*2*x^{21} = 6x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, and to get du/dx we use the power rule

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x^y = y*x^{y1}\] is the power rule

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can you finish solving the equation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok cause when he explained it in class i did not understand it at all.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0All right, how is it now?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok so for the power rule what do i plug in or do I have to plug anything in

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nah, 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\] \[\frac{d*ln(u)}{du} = \frac{1}{u}\] \[\frac{du}{dx} = 3*2*x^{21} = 6x\] \[u=3*x^2\] So you have \[\frac{1}{3*x^2} * 6x\] Simplify and you're done with that equation

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This maybe the hardest of the bunch because of the substitution, chain rule, and product rule

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, because substitution can be confusing because you introduce a new variable to differentiate on

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now (1.)\[y=e^x * lnx\]

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Heh, thanks, are you following along @Callisto ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y=e^x* lnx\] Here we use the product rule. We can think of this equation as the product of 2 functions of x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(f*g)' = f'*g + f*g'\] 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
 4 years ago
Best ResponseYou've already chosen the best response.2For 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, that is easier! @hooverst If you did it that way you don't have do substitution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It 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
 4 years ago
Best ResponseYou've already chosen the best response.0Nice one @Callisto @hooverst So, you think you can solve equation (1.) with the product rule?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Heh, you're right, man

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so callisto when you solved your equation for the second 1 what did you get for your anserw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks, getting a little carried away. Let me step away for a bit

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are we still working on the second equation? @hooverst ???

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in 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
 4 years ago
Best ResponseYou've already chosen the best response.0and either way the derivative is the same...

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, that is a derivative you need to memorize, it will be quite useful

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hehe, right, so, what is your opinion? ;p You can solve a problem many, many, different ways

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For you, how would you approach it now that you've seen how we would?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Definitely, that was an awesome application of intuition

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, and mathematical understanding

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i guess it comes down to preference because i personally would've used u'/u ... the method used earlier...

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2The 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.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would you multiply it.

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2Nope.... 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
 4 years ago
Best ResponseYou've already chosen the best response.0Ok is it d/dx(ln(x))=1/x

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

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok I get it. thanks Callisto
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