anonymous
  • anonymous
Can anyone break this down for me into steps? i need to find dy/dx by implicit differentiation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[2\sqrt{x}+\sqrt{y}=5\]
anonymous
  • anonymous
I am thinking it is \[1/4x^-1/2+1/2y^-1/2\]
anonymous
  • anonymous
those should be raised to the -1/2 but it doesn't look like it

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amistre64
  • amistre64
implicit is the same process as explicity; you just keep the dy and dx in until the end
amistre64
  • amistre64
2x^(1/2) + y^(1/2) = 5 ; derive everything with respect to x
amistre64
  • amistre64
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
amistre64
  • amistre64
now dx = 1 and dy=y' solve for y' 1/sqrt(x) + y' (1/2sqrt(y)) = 0
amistre64
  • amistre64
y' = [-1/sqrt(x)]/[1/2sqrt(y)] y ' = 2sqrt(y)/sqrt(x)
amistre64
  • amistre64
-2sqrt(x)/......
amistre64
  • amistre64
i lost the negative lol
anonymous
  • anonymous
OK trying to process this really quick
amistre64
  • amistre64
its confusing at first becasue we are used to discarding that dx.... keep all the bits in and adjust for them in the end
amistre64
  • amistre64
y = 2x dy = 2 dx right? dx=1 becasue it always changes the same with respect to itself
amistre64
  • amistre64
dy is just another term for y'
anonymous
  • anonymous
Alright, I think i almost with you. So when i break it down it's going dy/dx*f(x)
amistre64
  • amistre64
sorta, cant say im familiar with your notation there
amistre64
  • amistre64
lets take itby bits smaller pieces so you can see it work
anonymous
  • anonymous
well just the dirvitive of y/x times the starting function?
amistre64
  • amistre64
3y^2 what is the derivative of this with respect to x?
anonymous
  • anonymous
6y^1
amistre64
  • amistre64
you threw out the dy...how come?
anonymous
  • anonymous
well you don't need the raised to the power of one but
amistre64
  • amistre64
3y^2 has no "x" in it; does it?
anonymous
  • anonymous
no it doesn't
amistre64
  • amistre64
so the dy/dx that you derive doesnt go away does it?
amistre64
  • amistre64
the y with respect to x doesnt disappear does it?
anonymous
  • anonymous
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
amistre64
  • amistre64
its the change in y with respect to x: dy/dx
amistre64
  • amistre64
d (3y^2) dy (6y) ------- = --- dx dx
anonymous
  • anonymous
ok i'm with you
amistre64
  • amistre64
watch this one and Im sure your gonna slap yourself: d (5x^3) --------- = ? dx
amistre64
  • amistre64
d (5x^3) dx 15x^2 --------- = ---- dx dx
anonymous
  • anonymous
d (5x^3) dy (15x oh man! is that it?
amistre64
  • amistre64
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?
amistre64
  • amistre64
dx/dx = 1
amistre64
  • amistre64
we tend to through out the "dx/dx" because anything times 1 is itself... but its still there :)
amistre64
  • amistre64
ya with me?
anonymous
  • anonymous
yea i think i am getting it now. I don't know why this is giving me such a hard time.
anonymous
  • anonymous
so it's kinda like doing the point slope but with the change from f(x) to f'(x)
amistre64
  • amistre64
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 :)
anonymous
  • anonymous
not point slope but the slope finder
amistre64
  • amistre64
the product rule for xy: Dx(xy) = x'y + xy'
amistre64
  • amistre64
x' = 1 so its usually discarded
amistre64
  • amistre64
the power rule for y^5: y' 5y^4
anonymous
  • anonymous
so my dy/dx on that would be 5? because it's changed for a total of 5 in the coeficent?
amistre64
  • amistre64
your dy/dx inthat one is undertermined until you kow what to plug in for a value of y
amistre64
  • amistre64
but when you do, then you can determine dy/dx = y'
amistre64
  • amistre64
5xy^2 = z derive with respect to t?
amistre64
  • amistre64
5x2y y' + 5 x'y^2 = z'
amistre64
  • amistre64
there are no "t" values so all your derivative bits stay put
amistre64
  • amistre64
y' = dy/dt in this case x' = dx/dt z' = dz/dt
anonymous
  • anonymous
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)
amistre64
  • amistre64
the: 2sqrt(x) + sqrt(y) =5 one?
anonymous
  • anonymous
or\[2\sqrt{x}*1/2+1/2y^1/2\]
amistre64
  • amistre64
Dx(2sqrt(x)) + Dx(sqrt(y)) = Dx(5)
amistre64
  • amistre64
x' (1/sqrt(x)) + y' (1/2sqrt(y)) = 0
amistre64
  • amistre64
x' = dx/dx = 1 so we can ignore it.... or leave it in. solve for y'
amistre64
  • amistre64
there is no f'(x)*f(x) + dy/dx...... your mixing notations and confusing yoursoef :)
anonymous
  • anonymous
ok so it looks like you are just adding the dirivites inorder to solve this. am i hitting left field with that one?
amistre64
  • amistre64
\[\frac{dx}{dx}\frac{1}{\sqrt{x}} + \frac{dy}{dx}\frac{1}{2 \sqrt{y}} = 0\]
amistre64
  • amistre64
solve for dy/dx which is another notation for y'
anonymous
  • anonymous
Alright i think i am getting it now. Fantasitic answer as always thanks a ton
amistre64
  • amistre64
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
anonymous
  • anonymous
Yea i think i was reading into it more than it was really asking me to
amistre64
  • amistre64
y^2 = 5x^2 + 6x +3
amistre64
  • amistre64
\[\frac{dy}{dx}2y = \frac{dx}{dx}10x + \frac{dx}{dx}6 +\frac{dx}{dx}0\]
amistre64
  • amistre64
y' 2y = 10x +6
amistre64
  • amistre64
10x + 6 y' = --------- 2y
amistre64
  • amistre64
its just deriving like normal; just keep the bit in that you were taught to throw out lol
anonymous
  • anonymous
OH!!! so it's the change in y over the change in x
amistre64
  • amistre64
yes :)
anonymous
  • anonymous
i see what you did there
amistre64
  • amistre64
no you see what youve always done there ;)
anonymous
  • anonymous
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
amistre64
  • amistre64
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.
anonymous
  • anonymous
Well if they did things the easy way there would be a lot of math teachers out of a job.
amistre64
  • amistre64
lol maybe :)
anonymous
  • anonymous
Thanks for your time great explianition.
amistre64
  • amistre64
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
amistre64
  • amistre64
youre welcome :)
anonymous
  • anonymous
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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