Trisarahtops
  • Trisarahtops
find dy/dx for 4 – xy = y^3.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
SolomonZelman
  • SolomonZelman
Can you differentiate y•x?
Trisarahtops
  • Trisarahtops
that'd just be y
SolomonZelman
  • SolomonZelman
You are using the product rule, and the chain rule every time you take derivative of y, and that is you multiply times y'.

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SolomonZelman
  • SolomonZelman
What would be the derivative of \(\color{#000000 }{ \displaystyle x\cdot f(x) }\) ?
SolomonZelman
  • SolomonZelman
Use the product rule for that ...
SolomonZelman
  • SolomonZelman
The derivative of \(f(x)\) is \(f'(x)\) and the derivative of \(x\) is?
Trisarahtops
  • Trisarahtops
1
SolomonZelman
  • SolomonZelman
for \(x\), and for f(x) ?
Trisarahtops
  • Trisarahtops
no just x
SolomonZelman
  • SolomonZelman
Wait, the derivative of \(x\) with respect to x is 1. Right?
Trisarahtops
  • Trisarahtops
yes
SolomonZelman
  • SolomonZelman
And the derivative of \(f(x)\) is \(f'(x)\) ok?
Trisarahtops
  • Trisarahtops
right
SolomonZelman
  • SolomonZelman
By the product rule (for differentiable functions of x, F and G) d/dx [F•G] = F' • G + F • G' Ok?
Trisarahtops
  • Trisarahtops
yes
SolomonZelman
  • SolomonZelman
d/dx [x•f(x)] = ? (you tell me)
Trisarahtops
  • Trisarahtops
2f(x) ?
jim_thompson5910
  • jim_thompson5910
h(x) = g(x) * f(x) h(x) = x * f(x) so, g(x) = x g ' (x) = 1 Using the product rule, we get h ' (x) = g ' (x) * f(x) + g(x)*f ' (x) h ' (x) = _________ (fill in the blank)
SolomonZelman
  • SolomonZelman
You seem to struggle doing the product rule...
SolomonZelman
  • SolomonZelman
Can you differentiate \(xe^x\) ? (Knowing that the derivative of \(e^x\) is \(e^x\) and derivative of x is 1)
SolomonZelman
  • SolomonZelman
Product rule: d/dx (f•g) = f`g + fg`
SolomonZelman
  • SolomonZelman
Want some examples of product rule?
Trisarahtops
  • Trisarahtops
e^e ?
SolomonZelman
  • SolomonZelman
correct yourself please.
SolomonZelman
  • SolomonZelman
The derivative of \(e^x\) is just \(e^x\).
Trisarahtops
  • Trisarahtops
okay so now what?
SolomonZelman
  • SolomonZelman
Suppose I have the following function: \(\color{#000000 }{ \displaystyle h(x)=xe^x }\) Then the derivative of \(h(x)\) would be, \(\color{#000000 }{ \displaystyle \frac{d}{dx}~xe^x=\left(\frac{d}{dx} x \right) e^x+\left(\frac{d}{dx} e^x \right) x }\) The derivative of \(x\) (or \(x^1\)) is: \(\color{blue}{1}\cdot x^{\color{blue}{1}\color{red}{-1}}=1\cdot x^0=1\) (Knowing that \(y^0=1\) for all y besides y=0) The derivative of \(e^x\) is just \(e^x\). (If you like we can go through proving this via first principles of differentiation). So, we get; \(\color{#000000 }{ \displaystyle \frac{d}{dx}~(xe^x)=\left(\frac{d}{dx} x \right) e^x+\left(\frac{d}{dx} e^x \right) x }\) \(\color{#000000 }{ \displaystyle \frac{d}{dx}~xe^x=\left(1 \right) e^x+\left(e^x \right) x }\) \(\color{#000000 }{ \displaystyle \frac{d}{dx}~xe^x= e^x+\left(e^x \right) x }\)
SolomonZelman
  • SolomonZelman
If you see something unfamiliar or questionable in what I said, then please ask.
SolomonZelman
  • SolomonZelman
(take your time tho)
Trisarahtops
  • Trisarahtops
I think I understand but how do I use this to solve the problem?
mathmale
  • mathmale
Please note:\[4-xy=y^3\]
mathmale
  • mathmale
is an implicit function. Theoretically, it could be solved for y, but that'd be hard and not worth the trouble.
SolomonZelman
  • SolomonZelman
Did you get the product rule? Do you know the power rule (what I applied to differentiate the x)? Do you know the chain rule?
mathmale
  • mathmale
So you must jump in and take the derivative of each term with respect to x.
mathmale
  • mathmale
In other words, evaluate\[\frac{ d }{ dx}(4-xy=y^3)\]
mathmale
  • mathmale
what is \[\frac{ d }{ dx}4?\]
mathmale
  • mathmale
Hint: the derivative of a constant is always ___________
Trisarahtops
  • Trisarahtops
zero
mathmale
  • mathmale
Right. Now let's move on to
mathmale
  • mathmale
\[\frac{ d }{ dx }[-xy]\]
mathmale
  • mathmale
You could simplify matters by taking the neg. sign out.
Trisarahtops
  • Trisarahtops
-y
mathmale
  • mathmale
\[-\frac{ d }{ dx }xy\]
Trisarahtops
  • Trisarahtops
or do you want y?
mathmale
  • mathmale
solomon has gone over that with you already; please go back for a quick look at the product rule, and then at the derivative of xy with respect to x.
mathmale
  • mathmale
Again: xy is a product, and we want the derivative of this product, so "do you want y?" is irrelevant.
mathmale
  • mathmale
So, writing everything out formally, the derivative of xy is
mathmale
  • mathmale
...I'll let Solomon type that for us.
Trisarahtops
  • Trisarahtops
no i'm say the derivative of xy is y right?
SolomonZelman
  • SolomonZelman
Looks like there is a little lack of prerequisite knowledge... but we can fix that if you bear with us.... it might take time, but wouldn't you agree it is worth it?
SolomonZelman
  • SolomonZelman
Do you know the Power Rule? If you do please differentiate \(x^3\) for me.
mathmale
  • mathmale
Trisa: Sorry, no. Solomon: Please stick to the derivative of xy first, then worry about the power rule later.
mathmale
  • mathmale
Formally, the derivative with respect to x of xy is as follows:\[\frac{ d }{ dx }xy=x \frac{ dy }{ dx }+y \frac{ dx }{ dx }\]
mathmale
  • mathmale
This is the applicaTION OF THE product rule for differentiation to xy. Note that we are assuming that y is a function of x. That's why I had to say NO to your question earlier.
mathmale
  • mathmale
So, Tri-: can you simplify dx/dx?
Trisarahtops
  • Trisarahtops
1
mathmale
  • mathmale
Good! So, what do you have left?
Trisarahtops
  • Trisarahtops
just y
Trisarahtops
  • Trisarahtops
so now I simplify dy/dx?
mathmale
  • mathmale
Yes.\[\frac{ d }{ dx }xy=x \frac{ dy }{ dx }+y \]
Trisarahtops
  • Trisarahtops
d/dx xy = x y/2 +y
mathmale
  • mathmale
No. Please move on to the last term, which is y^3. What is the derivative of that?
mathmale
  • mathmale
I'm sorry, but x y/2 + y is not correct. Please compare it to my last previous equation, above.
mathmale
  • mathmale
I'm getting the results I have by applying the product rule to xy.
mathmale
  • mathmale
Try once more. The middle term is -xy. What is the derivative, with respect to x, of that?
Trisarahtops
  • Trisarahtops
didn't u just tell me what the derivative of xy was just now??
mathmale
  • mathmale
Yes. I want to be sure you agree with that before we move on.
mathmale
  • mathmale
\[\frac{ d }{ dx }xy=x \frac{ dy }{ dx }+y\]
Trisarahtops
  • Trisarahtops
x d/dx + y dx/dx
mathmale
  • mathmale
Now enclose that in parentheses and write a negative sign in front. Then you'll have the derivative of -xy with respect to x, and assuming that y is a function of x.
Trisarahtops
  • Trisarahtops
alright but how is dy/dx not simplified to y/x
mathmale
  • mathmale
\[\frac{ d }{ dx }[-xy]=-(x \frac{ dy }{ dx }+y)\]
mathmale
  • mathmale
I'm glad you asked that question. dy/dx cannot in any way be reduced. d is not a number, but an operator (command). dy/dx as a whole reads "the derivative of y with respect to x."
Trisarahtops
  • Trisarahtops
ha nvm lets just continue
Trisarahtops
  • Trisarahtops
so now I find the derivative of y^3?
mathmale
  • mathmale
The last term we have to deal with is y^3. Find the derivative of that with respect to x. Be certain to apply the power rule AND the chain rule.
Trisarahtops
  • Trisarahtops
3y^2
mathmale
  • mathmale
Great, but you haven't yet applied the chain rule. Remember, y is assumed to be a function of x, and thus you must differentiate y with respect to x.
mathmale
  • mathmale
So, the derivative of y^3 with respect to x is 3y^2 times what?
Trisarahtops
  • Trisarahtops
sooo zero?
mathmale
  • mathmale
The derivative of y with respect to x is denoted by dy/dx. You will see this again and again. Write the whole derivative of y^3 with respect to x here:
mathmale
  • mathmale
Everything you need to do that is included in the most recent 3 entries, above.
mathmale
  • mathmale
compare this to
mathmale
  • mathmale
\[3y^2\frac{ dy }{ dx }\]
mathmale
  • mathmale
sorry it's hard for you to understand this right now, but dy/dx is NOT equal to 1. Instead, dy/dx is a LABEL (like your name is a label) for "the derivative of y with respect to x."
mathmale
  • mathmale
To get 3y^2, we used the power rule. We remember that y is assumed to be a function of x, so we follow the chain rule and write dy/dx immediately after 3y^2.
mathmale
  • mathmale
In summary:\[\frac{ d }{ dx }4=0\]
mathmale
  • mathmale
and ...
mathmale
  • mathmale
\[-\frac{ d }{ dx }xy =-[x \frac{ dy }{ dx }+y]\]
mathmale
  • mathmale
... and ...\[\frac{ d }{ dx }y^3=3y^2\frac{ dy }{ dx }\]
mathmale
  • mathmale
I'm afraid you'll have to take my word for this for the time being. You can ask all the questions you want, perhaps tomorrow.
mathmale
  • mathmale
So, put all 3 of these derivatives together to obtain the derivative of the implicit function 4-xy=y^3.
mathmale
  • mathmale
Could you do that now, please? Don't bother with \[\frac{ d }{ dx}4,\]
mathmale
  • mathmale
because it's zero. What's left?
Trisarahtops
  • Trisarahtops
-[x dy/dx+ y] = 3y^2 dy/dx
mathmale
  • mathmale
You are now writing an equation with the derivative of -xy on the left of the = sign and the deriv. of y^3 on the right. Yes, y ou have that correct.
mathmale
  • mathmale
The last step of the process of finding the derivative is to SOLVE FOR dy/dx. Once again, dy/dx reads "the derivative of y with respect to x" and is not 1 and is not 0.
mathmale
  • mathmale
There are 3 terms in the equation you've just written. 2 have dy/dx in them, and 1 does not. Can you agree with that?
Trisarahtops
  • Trisarahtops
yes
mathmale
  • mathmale
First, I'll type the 2 terms that have dy/dx in them on the left. Then I'll put the one term that does not have dy/dx on the right.
mathmale
  • mathmale
-x(dy/dx) - y = 3y^2(dy/dx)
mathmale
  • mathmale
(dy/dx terms on the left: see the following:)
mathmale
  • mathmale
\[-x \frac{ dy }{ dx }-3y^2\frac{ dy }{ dx }=y\]
mathmale
  • mathmale
We must solve this for (dy/dx).
mathmale
  • mathmale
Factoring out (dy/dx), \[\frac{ dy }{ dx }[-x-3y^2]=y\]
mathmale
  • mathmale
What's next? Reminder: we are solving this equation for dy/dx.
mathmale
  • mathmale
OpenStudy tells me that you're "just looking around" at the moment. What remains to be done in finding the derivative dy/dx is to solve the very last equation (above) for dy/dx. Message me (or reply here) if you have further questions about this problem. I have to get off the 'Net now. Thanks for your perseverance and patience.
mathmale
  • mathmale
Good night.

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