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NathanJHW
 one year ago
If y^5 +3x^2y^2 +5x^4 = 49 , then dy/dx at the point (1, 2) is:
NathanJHW
 one year ago
If y^5 +3x^2y^2 +5x^4 = 49 , then dy/dx at the point (1, 2) is:

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NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0−1 11/23 −23/11 −10/3 0

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\rm y^5 +3x^2y^2 +5x^4 = 49\]So looks like you'll want to differentiate implicitly. Having trouble with that step? :o

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2We're taking our derivative with respect to x,\[\Large\rm (y^5)'\ne5y^4\]So whenever you differentiate a nonx thing, you end up with this dy/dx or y' term by the chain rule,\[\Large\rm (y^5)'=5y^4y'\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\rm \color{orangered}{(y^5)'} +(3x^2y^2)' +(5x^4)' = (49)'\]\[\Large\rm \color{orangered}{5y^4y'} +(3x^2y^2)' +(5x^4)' = (49)'\]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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah that makes sense to me.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So what do you get for the orange stuff? :o

NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0Do you mean what did I get when I did the work or do I solve your work now?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0We can continue with your work, I was no where close so it's all wrong.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So 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
 one year ago
Best ResponseYou've already chosen the best response.0((dy/dx)(5y^4y)) +((dy/dx)(3x^2y^2))+ ((d/dx)(5x^4))=(d/dx)(49)

NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0Is that the correct setup @zepdrix

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2If 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
 one year ago
Best ResponseYou've already chosen the best response.2I think you're having a little trouble with your chain rule... or something.. hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\rm \frac{d}{dx}5x^4=5\cdot4x^3\cdot \frac{d}{dx}x=20x^3\cdot \frac{dx}{dx}=20x^3\cdot 1\]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
 one year ago
Best ResponseYou've already chosen the best response.2Alternatively 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433291724073:dwYou'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.

NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0So how does this lead into the answer?

NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0I understand how to get this, but where do we go next?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Well we've gotta get these terms set up correctly. Your middle derivative is looking strange.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2First 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
 one year ago
Best ResponseYou've already chosen the best response.2Right 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
 one year ago
Best ResponseYou've already chosen the best response.2Middle 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
 one year ago
Best ResponseYou've already chosen the best response.2And 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm 5y^4\frac{dy}{dx} +\color{orangered}{y^2(6x)}+\color{royalblue}{3x^2\frac{d}{dx}(y^2)} +5\cdot4x^3 = 0\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2And 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
 one year ago
Best ResponseYou've already chosen the best response.2And 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm 5(2)^4\frac{dy}{dx} +(2)^2(6(1))+3(1)^2(2(2))\frac{dy}{dx} +5\cdot4(1)^3 = 0\]Something like that, ya? I probably used more brackets than I needed to _ Anyway, then you solve for dy/dx.
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