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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

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

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?

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?

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)

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

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 :)

\[\frac{dx}{dx}\frac{1}{\sqrt{x}} + \frac{dy}{dx}\frac{1}{2 \sqrt{y}} = 0\]

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

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 ;)

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.

youre welcome :)