Question

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

Answer 1

@perl

Answer 2

−1
11/23
−23/11
−10/3
0

Answer 3

@Hero

Answer 4

@zepdrix

Answer 5

\[\Large\rm y^5 +3x^2y^2 +5x^4 = 49\]So looks like you'll want to differentiate implicitly.
Having trouble with that step? :o

Answer 6

Yes I am.

Answer 7

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,\[\Large\rm (y^5)'=5y^4y'\]

Answer 8

\[\Large\rm \color{orangered}{(y^5)'} +(3x^2y^2)' +(5x^4)' = (49)'\]\[\Large\rm \color{orangered}{5y^4y'} +(3x^2y^2)' +(5x^4)' = (49)'\]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

Answer 9

Yeah that makes sense to me.

Answer 10

So what do you get for the orange stuff? :o

Answer 11

Do you mean what did I get when I did the work or do I solve your work now?

Answer 12

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

Answer 13

We can continue with your work, I was no where close so it's all wrong.

Answer 14

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)

Answer 15

((dy/dx)(5y^4y)) +((dy/dx)(3x^2y^2))+ ((d/dx)(5x^4))=(d/dx)(49)

Answer 16

Is that the correct setup @zepdrix

Answer 17

hmm

Answer 18

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:\[\Large\rm \color{orangered}{5y^4\frac{dy}{dx}} +\frac{d}{dx}(3x^2y^2) +\frac{d}{dx}(5x^4) = \frac{d}{dx}(49)\]

Answer 19

I think you're having a little trouble with your chain rule...
or something.. hmm

Answer 20

\[\Large\rm \frac{d}{dx}5x^4=5\cdot4x^3\cdot \frac{d}{dx}x=20x^3\cdot \frac{dx}{dx}=20x^3\cdot 1\]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.

Answer 21

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.

Answer 22

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.