anonymous
  • anonymous
y=e^xlnx y=ln(3x^2) y=lnx/x^2+1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
what's the question? which equation is nicest?
anonymous
  • anonymous
What's the goal? Solve for x in this system of equations?
anonymous
  • anonymous
heh, I like the first one ;)

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

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