NathanJHW
  • NathanJHW
If y^5 +3x^2y^2 +5x^4 = 49 , then dy/dx at the point (-1, 2) is:
Mathematics
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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!
NathanJHW
  • NathanJHW
@perl
NathanJHW
  • NathanJHW
−1 11/23 −23/11 −10/3 0
NathanJHW
  • NathanJHW
@Hero

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NathanJHW
  • NathanJHW
@zepdrix
zepdrix
  • zepdrix
\[\Large\rm y^5 +3x^2y^2 +5x^4 = 49\]So looks like you'll want to differentiate implicitly. Having trouble with that step? :o
NathanJHW
  • NathanJHW
Yes I am.
zepdrix
  • zepdrix
We're taking our derivative with respect to x,\[\Large\rm (y^5)'\ne5y^4\]So whenever you differentiate a non-x thing, you end up with this dy/dx or y' term by the chain rule,\[\Large\rm (y^5)'=5y^4y'\]
zepdrix
  • zepdrix
\[\Large\rm \color{orangered}{(y^5)'} +(3x^2y^2)' +(5x^4)' = (49)'\]\[\Large\rm \color{orangered}{5y^4y'} +(3x^2y^2)' +(5x^4)' = (49)'\]The middle part is a little trickier. You have two variable things being multiplied together, so we'll need to apply our product rule. It doesn't matter which you group the 3 with, just choose one of them.\[\Large\rm 5y^4y' +\color{orangered}{(3x^2y^2)'} +(5x^4)' = (49)'\]\[\Large\rm 5y^4y' +\color{orangered}{(3x^2)'y^2+3x^2(y^2)'} +(5x^4)' = (49)'\]That product rule setup make sense? :o
NathanJHW
  • NathanJHW
Yeah that makes sense to me.
zepdrix
  • zepdrix
So what do you get for the orange stuff? :o
NathanJHW
  • NathanJHW
Do you mean what did I get when I did the work or do I solve your work now?
zepdrix
  • zepdrix
Oh you already attempt this problem? :) Yah either is fine, we can continue doing this, or you can tell me what you got doing your way and I can check it. either is fine
NathanJHW
  • NathanJHW
We can continue with your work, I was no where close so it's all wrong.
zepdrix
  • zepdrix
So let's just worry about the orange part right now, what do you get when you take those derivatives? (In one of them, you're taking derivative of y, so don't forget about your dy/dx)
NathanJHW
  • NathanJHW
((dy/dx)(5y^4y)) +((dy/dx)(3x^2y^2))+ ((d/dx)(5x^4))=(d/dx)(49)
NathanJHW
  • NathanJHW
Is that the correct setup @zepdrix
zepdrix
  • zepdrix
hmm
zepdrix
  • zepdrix
If you don't like the primes, fine we can use the differentials instead. So we have:\[\Large\rm \frac{d}{dx}(y^5 +3x^2y^2 +5x^4) = \frac{d}{dx}(49)\] \[\Large\rm \frac{d}{dx}(y^5) +\frac{d}{dx}(3x^2y^2) +\frac{d}{dx}(5x^4) = \frac{d}{dx}(49)\]First term gives you:\[\Large\rm \color{orangered}{5y^4\frac{dy}{dx}} +\frac{d}{dx}(3x^2y^2) +\frac{d}{dx}(5x^4) = \frac{d}{dx}(49)\]
zepdrix
  • zepdrix
I think you're having a little trouble with your chain rule... or something.. hmm
zepdrix
  • zepdrix
\[\Large\rm \frac{d}{dx}5x^4=5\cdot4x^3\cdot \frac{d}{dx}x=20x^3\cdot \frac{dx}{dx}=20x^3\cdot 1\]With your third term here, keep in mind that we're differentiating with respect to x, so you don't end up with a dy/dx on the last term, you actually end up with a "dx/dx" which is insignificant.
zepdrix
  • zepdrix
Alternatively what I did with the y term was,\[\Large\rm \frac{d}{dx}y^5=5y^4\cdot\frac{d}{dx}y=5y^4\cdot \frac{dy}{dx}\]chain rule again, but now we end up with something that actually has meaning, this dy/dx.
zepdrix
  • zepdrix
|dw:1433291724073:dw|You're always applying your chain rule, even just with the x's. We often don't show that work because it has no significance. If we're differentiating x with respect to x, we don't end up with any extra stuff.
zepdrix
  • zepdrix
Too confusing? :O
NathanJHW
  • NathanJHW
So how does this lead into the answer?
NathanJHW
  • NathanJHW
I understand how to get this, but where do we go next?
zepdrix
  • zepdrix
Well we've gotta get these terms set up correctly. Your middle derivative is looking strange.
zepdrix
  • zepdrix
First term gave us this,\[\Large\rm \color{orangered}{5y^4\frac{dy}{dx}} +\frac{d}{dx}(3x^2y^2) +\frac{d}{dx}(5x^4) = \frac{d}{dx}(49)\]Third term will give us this, no dy/dx on this guy,\[\Large\rm 5y^4\frac{dy}{dx} +\frac{d}{dx}(3x^2y^2) +\color{orangered}{5\cdot4x^3} = \frac{d}{dx}(49)\]
zepdrix
  • zepdrix
Right side will give us this:\[\Large\rm 5y^4\frac{dy}{dx} +\frac{d}{dx}(3x^2y^2) +5\cdot4x^3 = \color{orangered}{0}\]
zepdrix
  • zepdrix
Middle term we have to setup our product rule before we apply any differentiation rules like power rule or chain rule. \[\large\rm 5y^4\frac{dy}{dx} +\color{royalblue}{y^2\frac{d}{dx}(3x^2)+3x^2\frac{d}{dx}(y^2)} +5\cdot4x^3 = 0\]
zepdrix
  • zepdrix
And then apply your rules to each term individually, you shouldn't end up with a dy/dx on the whole group, i was a little confused by that.
zepdrix
  • zepdrix
\[\large\rm 5y^4\frac{dy}{dx} +\color{orangered}{y^2(6x)}+\color{royalblue}{3x^2\frac{d}{dx}(y^2)} +5\cdot4x^3 = 0\]
zepdrix
  • zepdrix
And the other part gives you,\[\large\rm 5y^4\frac{dy}{dx} +y^2(6x)+\color{orangered}{3x^2(2y)\frac{dy}{dx}} +5\cdot4x^3 = 0\]
zepdrix
  • zepdrix
And then you want to evaluate your derivative function dy/dx at the point (-1,2). So plug in `-1 for all the x` and `2 for all the y`.
zepdrix
  • zepdrix
\[\large\rm 5(2)^4\frac{dy}{dx} +(2)^2(6(-1))+3(-1)^2(2(2))\frac{dy}{dx} +5\cdot4(-1)^3 = 0\]Something like that, ya? I probably used more brackets than I needed to -_- Anyway, then you solve for dy/dx.

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