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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2\sqrt{x}+\sqrt{y}=5\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am thinking it is \[1/4x^1/2+1/2y^1/2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0those should be raised to the 1/2 but it doesn't look like it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0implicit is the same process as explicity; you just keep the dy and dx in until the end

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02x^(1/2) + y^(1/2) = 5 ; derive everything with respect to x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Dx(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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now dx = 1 and dy=y' solve for y' 1/sqrt(x) + y' (1/2sqrt(y)) = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y' = [1/sqrt(x)]/[1/2sqrt(y)] y ' = 2sqrt(y)/sqrt(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i lost the negative lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK trying to process this really quick

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its confusing at first becasue we are used to discarding that dx.... keep all the bits in and adjust for them in the end

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = 2x dy = 2 dx right? dx=1 becasue it always changes the same with respect to itself

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0dy is just another term for y'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, I think i almost with you. So when i break it down it's going dy/dx*f(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sorta, cant say im familiar with your notation there

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets take itby bits smaller pieces so you can see it work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well just the dirvitive of y/x times the starting function?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.03y^2 what is the derivative of this with respect to x?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you threw out the dy...how come?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well you don't need the raised to the power of one but

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.03y^2 has no "x" in it; does it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so the dy/dx that you derive doesnt go away does it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the y with respect to x doesnt disappear does it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its the change in y with respect to x: dy/dx

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0d (3y^2) dy (6y)  =  dx dx

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0watch this one and Im sure your gonna slap yourself: d (5x^3)  = ? dx

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0d (5x^3) dx 15x^2  =  dx dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0d (5x^3) dy (15x oh man! is that it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0tell 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we tend to through out the "dx/dx" because anything times 1 is itself... but its still there :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i think i am getting it now. I don't know why this is giving me such a hard time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it's kinda like doing the point slope but with the change from f(x) to f'(x)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its 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 :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not point slope but the slope finder

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the product rule for xy: Dx(xy) = x'y + xy'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x' = 1 so its usually discarded

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the power rule for y^5: y' 5y^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so my dy/dx on that would be 5? because it's changed for a total of 5 in the coeficent?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0your dy/dx inthat one is undertermined until you kow what to plug in for a value of y

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but when you do, then you can determine dy/dx = y'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05xy^2 = z derive with respect to t?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05x2y y' + 5 x'y^2 = z'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0there are no "t" values so all your derivative bits stay put

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y' = dy/dt in this case x' = dx/dt z' = dz/dt

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the: 2sqrt(x) + sqrt(y) =5 one?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or\[2\sqrt{x}*1/2+1/2y^1/2\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Dx(2sqrt(x)) + Dx(sqrt(y)) = Dx(5)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x' (1/sqrt(x)) + y' (1/2sqrt(y)) = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x' = dx/dx = 1 so we can ignore it.... or leave it in. solve for y'

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0there is no f'(x)*f(x) + dy/dx...... your mixing notations and confusing yoursoef :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so it looks like you are just adding the dirivites inorder to solve this. am i hitting left field with that one?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dx}{dx}\frac{1}{\sqrt{x}} + \frac{dy}{dx}\frac{1}{2 \sqrt{y}} = 0\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0solve for dy/dx which is another notation for y'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright i think i am getting it now. Fantasitic answer as always thanks a ton

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yea i think i was reading into it more than it was really asking me to

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{dy}{dx}2y = \frac{dx}{dx}10x + \frac{dx}{dx}6 +\frac{dx}{dx}0\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.010x + 6 y' =  2y

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its just deriving like normal; just keep the bit in that you were taught to throw out lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OH!!! so it's the change in y over the change in x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i see what you did there

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0no you see what youve always done there ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yea it almost seems like it would have been eaiser to start off with this way instead of having to go back and relearn it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0exactly; 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well if they did things the easy way there would be a lot of math teachers out of a job.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks for your time great explianition.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea 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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