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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

−1 11/23 −23/11 −10/3 0

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[\Large\rm y^5 +3x^2y^2 +5x^4 = 49\]So looks like you'll want to differentiate implicitly. Having trouble with that step? :o
Yes I am.
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,\[\Large\rm (y^5)'=5y^4y'\]
\[\Large\rm \color{orangered}{(y^5)'} +(3x^2y^2)' +(5x^4)' = (49)'\]\[\Large\rm \color{orangered}{5y^4y'} +(3x^2y^2)' +(5x^4)' = (49)'\]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
Yeah that makes sense to me.
So what do you get for the orange stuff? :o
Do you mean what did I get when I did the work or do I solve your work now?
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
We can continue with your work, I was no where close so it's all wrong.
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)
((dy/dx)(5y^4y)) +((dy/dx)(3x^2y^2))+ ((d/dx)(5x^4))=(d/dx)(49)
Is that the correct setup @zepdrix
hmm
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:\[\Large\rm \color{orangered}{5y^4\frac{dy}{dx}} +\frac{d}{dx}(3x^2y^2) +\frac{d}{dx}(5x^4) = \frac{d}{dx}(49)\]
I think you're having a little trouble with your chain rule... or something.. hmm
\[\Large\rm \frac{d}{dx}5x^4=5\cdot4x^3\cdot \frac{d}{dx}x=20x^3\cdot \frac{dx}{dx}=20x^3\cdot 1\]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.
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.
|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.
Too confusing? :O
So how does this lead into the answer?
I understand how to get this, but where do we go next?
Well we've gotta get these terms set up correctly. Your middle derivative is looking strange.
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,\[\Large\rm 5y^4\frac{dy}{dx} +\frac{d}{dx}(3x^2y^2) +\color{orangered}{5\cdot4x^3} = \frac{d}{dx}(49)\]
Right side will give us this:\[\Large\rm 5y^4\frac{dy}{dx} +\frac{d}{dx}(3x^2y^2) +5\cdot4x^3 = \color{orangered}{0}\]
Middle term we have to setup our product rule before we apply any differentiation rules like power rule or chain rule. \[\large\rm 5y^4\frac{dy}{dx} +\color{royalblue}{y^2\frac{d}{dx}(3x^2)+3x^2\frac{d}{dx}(y^2)} +5\cdot4x^3 = 0\]
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.
\[\large\rm 5y^4\frac{dy}{dx} +\color{orangered}{y^2(6x)}+\color{royalblue}{3x^2\frac{d}{dx}(y^2)} +5\cdot4x^3 = 0\]
And the other part gives you,\[\large\rm 5y^4\frac{dy}{dx} +y^2(6x)+\color{orangered}{3x^2(2y)\frac{dy}{dx}} +5\cdot4x^3 = 0\]
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`.
\[\large\rm 5(2)^4\frac{dy}{dx} +(2)^2(6(-1))+3(-1)^2(2(2))\frac{dy}{dx} +5\cdot4(-1)^3 = 0\]Something like that, ya? I probably used more brackets than I needed to -_- Anyway, then you solve for dy/dx.

Not the answer you are looking for?

Search for more explanations.

Ask your own question