Question

y=e^xlnx

y=ln(3x^2)

y=lnx/x^2+1

Answer 1

what's the question? which equation is nicest?

Answer 2

What's the goal? Solve for x in this system of equations?

Answer 3

heh, I like the first one ;)

Answer 4

me too! :)

Answer 5

each one is a different question its says compute the derivatives of each function

Answer 6

Oh, OK, are you familiar with (1.) product rule and (2.) quotient rule and the (3.) power rule?

Answer 7

yeah a little

Answer 8

(1.) \[y=e^x*ln(x)\]

(2.) \[y=ln(3x^2)\]

(3.) \[y=\frac{ln(x)}{x^2}+1\]

Answer 9

And you know chain rule? a little, right? ;)

Answer 10

the 1st equation is \[y=e ^{xlnx}\]

Answer 11

yes

Answer 12

OK, well, let's work with equation 2, for me that is the easiest one. We are differentiating with respect to y, right?

Answer 13

yes

Answer 14

Cool, \[y=ln(3*x^2)\] We will use the Chain Rule here

Answer 15

\[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\]
\[u = 3*x^2\]and \[\frac{d*ln(u)}{du} = \frac{1}{u}\]
\[\frac{du}{dx} = 3*2*x^{2-1} = 6x\]

Answer 16

Yes, and to get du/dx we use the power rule

Answer 17

\[x^y = y*x^{y-1}\] is the power rule

Answer 18

Can you finish solving the equation?

Answer 19

ok cause when he explained it in class i did not understand it at all.

Answer 20

All right, how is it now?

Answer 21

ok so for the power rule what do i plug in or do I have to plug anything in

Answer 22

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

Answer 23

\[\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}\]
\[\frac{d*ln(u)}{du} = \frac{1}{u}\]
\[\frac{du}{dx} = 3*2*x^{2-1} = 6x\]
\[u=3*x^2\]
So you have \[\frac{1}{3*x^2} * 6x\]
Simplify and you're done with that equation

Answer 24

O ok I didn't know whether what I had was right but thats what I just got to.

Answer 25

This maybe the hardest of the bunch because of the substitution, chain rule, and product rule

Answer 26

yea It was

Answer 27

Well, because substitution can be confusing because you introduce a new variable to differentiate on

Answer 28

*or with respect to

Answer 29

O ok

Answer 30

Now (1.)\[y=e^x * lnx\]

Answer 31

Excuse ... me.... are we finding dy/dx or dx/dy?

Answer 32

dy/dx I believe

Answer 33

Thank you. Please continue....

Answer 34

Heh, thanks, are you following along @Callisto ?

Answer 35

\[y=e^x* lnx\]
Here we use the product rule. We can think of this equation as the product of 2 functions of x

Answer 36

\[(f*g)' = f'*g + f*g'\]
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

Answer 37

For the second one, I was thinking in this way..
\[y=ln(3x^2)=ln3+ lnx^2 = ln3+2lnx\]\[\frac{dy}{dx}=\frac{d}{dx}2lnx = ...\]

Answer 38

Yes, that is easier! @hooverst If you did it that way you don't have do substitution

Answer 39

Ok dpflan

Answer 40

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

Answer 41

*to find

Answer 42

Nice one @Callisto

@hooverst So, you think you can solve equation (1.) with the product rule?

Answer 43

hold on... i think @hooverst clarified that the first equation is:
\(\large y=e^{xlnx} \) , not \(\large y=e^{x}\cdot lnx \)

Answer 44

Heh, you're right, man

Answer 45

so callisto when you solved your equation for the second 1 what did you get for your anserw

Answer 46

Thanks, getting a little carried away. Let me step away for a bit

Answer 47

ur doing fine....:)

Answer 48

@hooverst What did you get ??

Answer 49

is the anserw \[y=\ln(3x^2)=1\div 3x^2\]

Answer 50

Not really...
what is \(\frac{d}{dx}lnx\) ?

Answer 51

im not sure

Answer 52

are we still working on the second equation? @hooverst ???

Answer 53

because you can do this many different ways... but in the end, the derivative should be the same...

Answer 54

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 55

answer as in the derivative? you and @dpflan , worked it out to \(\large \frac{6x}{3x^2} \) simplified to....???

Answer 56

in the case with @Callisto , her method was to simplify the function first:
\(\large y=ln(3x^2)=ln3+2lnx \)
so
\(\large y'=[ln3]'+[2lnx]'= \)
...???

Answer 57

and either way the derivative is the same...

Answer 58

The key point is you need to know what \(\frac{d}{dx}lnx\) is.

Answer 59

Yes, that is a derivative you need to memorize, it will be quite useful

Answer 60

Ok so many of you have said had your own opinion about the equation so which one is the right 1.

Answer 61

Hehe, right, so, what is your opinion? ;p

You can solve a problem many, many, different ways

Answer 62

For you, how would you approach it now that you've seen how we would?

Answer 63

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.

Answer 64

Ok Think the one Callisto gave me is more simple for #2

Answer 65

Definitely, that was an awesome application of intuition

Answer 66

well, and mathematical understanding

Answer 67

i guess it comes down to preference because i personally would've used u'/u ... the method used earlier...

Answer 68

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.

Answer 69

ok

Answer 70

May I ask you again- what is d/dx ( lnx) ?

Answer 71

would you multiply it.

Answer 72

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.

Answer 73

Ok is it d/dx(ln(x))=1/x

Answer 74

Yes.
dy/dx = d/dx (ln3 + lnx^2) = d/dx (2lnx) = 2 d/dx(lnx) = ...?

Answer 75

is it 2/x or just 2x Idon't know if its right

Answer 76

Which one do you think? (i) 2/x (ii) 2x

Answer 77

2/x

Answer 78

Yes. That's correct. Any questions?

Answer 79

just 1 when we started simplifying the equation where did you get the 2 from?

Answer 80

\[lnx^a = a\ln x\]

Answer 81

Ok I get it. thanks Callisto

Answer 82

Welcome.