Can anyone break this down for me into steps? i need to find dy/dx by implicit differentiation?

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Can anyone break this down for me into steps? i need to find dy/dx by implicit differentiation?

Mathematics
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\[2\sqrt{x}+\sqrt{y}=5\]
I am thinking it is \[1/4x^-1/2+1/2y^-1/2\]
those should be raised to the -1/2 but it doesn't look like it

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implicit is the same process as explicity; you just keep the dy and dx in until the end
2x^(1/2) + y^(1/2) = 5 ; derive everything with respect to x
Dx(2x^(1/2)) = 1/sqrt(x) dx Dx(y^(1/2)) = 1/2sqrt(y) dy Dx(5) = 0 ....d5 is useless lol so it jsut 0
now dx = 1 and dy=y' solve for y' 1/sqrt(x) + y' (1/2sqrt(y)) = 0
y' = [-1/sqrt(x)]/[1/2sqrt(y)] y ' = 2sqrt(y)/sqrt(x)
-2sqrt(x)/......
i lost the negative lol
OK trying to process this really quick
its confusing at first becasue we are used to discarding that dx.... keep all the bits in and adjust for them in the end
y = 2x dy = 2 dx right? dx=1 becasue it always changes the same with respect to itself
dy is just another term for y'
Alright, I think i almost with you. So when i break it down it's going dy/dx*f(x)
sorta, cant say im familiar with your notation there
lets take itby bits smaller pieces so you can see it work
well just the dirvitive of y/x times the starting function?
3y^2 what is the derivative of this with respect to x?
6y^1
you threw out the dy...how come?
well you don't need the raised to the power of one but
3y^2 has no "x" in it; does it?
no it doesn't
so the dy/dx that you derive doesnt go away does it?
the y with respect to x doesnt disappear does it?
Ok maybe this is where i am going wrong when i look at dy/dx i think of a fraction of the dirvitive of y over the dirvitive of x
its the change in y with respect to x: dy/dx
d (3y^2) dy (6y) ------- = --- dx dx
ok i'm with you
watch this one and Im sure your gonna slap yourself: d (5x^3) --------- = ? dx
d (5x^3) dx 15x^2 --------- = ---- dx dx
d (5x^3) dy (15x oh man! is that it?
tell me; when x moves by any amount; x moves in respect to itself by the same amount right: lets say you are x and you move 5 feet to the right; how far have you moved?
dx/dx = 1
we tend to through out the "dx/dx" because anything times 1 is itself... but its still there :)
ya with me?
yea i think i am getting it now. I don't know why this is giving me such a hard time.
so it's kinda like doing the point slope but with the change from f(x) to f'(x)
its difficult becasue you are used to throwing away the "dx/dx" and keeping the dy/dx on the other side. Implicit is the same thing, but y aint by itself :)
not point slope but the slope finder
the product rule for xy: Dx(xy) = x'y + xy'
x' = 1 so its usually discarded
the power rule for y^5: y' 5y^4
so my dy/dx on that would be 5? because it's changed for a total of 5 in the coeficent?
your dy/dx inthat one is undertermined until you kow what to plug in for a value of y
but when you do, then you can determine dy/dx = y'
5xy^2 = z derive with respect to t?
5x2y y' + 5 x'y^2 = z'
there are no "t" values so all your derivative bits stay put
y' = dy/dt in this case x' = dx/dt z' = dz/dt
ok i think i am starting to get it so for my the equation i gave earlier it's going to work out to being f(x)+f'(x) +dy/dx+f'(y)
the: 2sqrt(x) + sqrt(y) =5 one?
or\[2\sqrt{x}*1/2+1/2y^1/2\]
Dx(2sqrt(x)) + Dx(sqrt(y)) = Dx(5)
x' (1/sqrt(x)) + y' (1/2sqrt(y)) = 0
x' = dx/dx = 1 so we can ignore it.... or leave it in. solve for y'
there is no f'(x)*f(x) + dy/dx...... your mixing notations and confusing yoursoef :)
ok so it looks like you are just adding the dirivites inorder to solve this. am i hitting left field with that one?
\[\frac{dx}{dx}\frac{1}{\sqrt{x}} + \frac{dy}{dx}\frac{1}{2 \sqrt{y}} = 0\]
solve for dy/dx which is another notation for y'
Alright i think i am getting it now. Fantasitic answer as always thanks a ton
if this was all in the variables for x youd hav eno problem finding the derivative :) because you were taught to throw out that dx/dx
Yea i think i was reading into it more than it was really asking me to
y^2 = 5x^2 + 6x +3
\[\frac{dy}{dx}2y = \frac{dx}{dx}10x + \frac{dx}{dx}6 +\frac{dx}{dx}0\]
y' 2y = 10x +6
10x + 6 y' = --------- 2y
its just deriving like normal; just keep the bit in that you were taught to throw out lol
OH!!! so it's the change in y over the change in x
yes :)
i see what you did there
no you see what youve always done there ;)
Yea it almost seems like it would have been eaiser to start off with this way instead of having to go back and relearn it
exactly; I dont know wy most of math is taught backwards, giving you exceptions to the rules instead of letting you learn the rules as they are.
Well if they did things the easy way there would be a lot of math teachers out of a job.
lol maybe :)
Thanks for your time great explianition.
its like: heres a shortcut way of shortcut say thats only going to be helpful for about 1% of the math that is going to be done; learn it, memorize it and then forget it later on when you realize that its useless lol
youre welcome :)
yea thats pretty much how my math career seems to have gone so far. It seems to me that it would be better to learn a process that works 100% of the time then go back and learn all the shortcuts that will speed up your calculation time on equations. However, i have always been warned "becareful what you wish for" that might open open a can of worms that shouldn't be touched.

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