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

- anonymous

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

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- anonymous

\[2\sqrt{x}+\sqrt{y}=5\]

- anonymous

I am thinking it is \[1/4x^-1/2+1/2y^-1/2\]

- anonymous

those should be raised to the -1/2 but it doesn't look like it

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## More answers

- amistre64

implicit is the same process as explicity; you just keep the dy and dx in until the end

- amistre64

2x^(1/2) + y^(1/2) = 5 ; derive everything with respect to x

- 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

now dx = 1 and dy=y' solve for y'
1/sqrt(x) + y' (1/2sqrt(y)) = 0

- amistre64

y' = [-1/sqrt(x)]/[1/2sqrt(y)]
y ' = 2sqrt(y)/sqrt(x)

- amistre64

-2sqrt(x)/......

- amistre64

i lost the negative lol

- anonymous

OK trying to process this really quick

- 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

y = 2x
dy = 2 dx right? dx=1 becasue it always changes the same with respect to itself

- amistre64

dy is just another term for y'

- anonymous

Alright, I think i almost with you. So when i break it down it's going dy/dx*f(x)

- amistre64

sorta, cant say im familiar with your notation there

- amistre64

lets take itby bits smaller pieces so you can see it work

- anonymous

well just the dirvitive of y/x times the starting function?

- amistre64

3y^2 what is the derivative of this with respect to x?

- anonymous

6y^1

- amistre64

you threw out the dy...how come?

- anonymous

well you don't need the raised to the power of one but

- amistre64

3y^2 has no "x" in it; does it?

- anonymous

no it doesn't

- amistre64

so the dy/dx that you derive doesnt go away does it?

- amistre64

the y with respect to x doesnt disappear does it?

- 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

its the change in y with respect to x: dy/dx

- amistre64

d (3y^2) dy (6y)
------- = ---
dx dx

- anonymous

ok i'm with you

- amistre64

watch this one and Im sure your gonna slap yourself:
d (5x^3)
--------- = ?
dx

- amistre64

d (5x^3) dx 15x^2
--------- = ----
dx dx

- anonymous

d (5x^3) dy (15x
oh man! is that it?

- 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

dx/dx = 1

- amistre64

we tend to through out the "dx/dx" because anything times 1 is itself... but its still there :)

- amistre64

ya with me?

- anonymous

yea i think i am getting it now. I don't know why this is giving me such a hard time.

- anonymous

so it's kinda like doing the point slope but with the change from f(x) to f'(x)

- 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

not point slope but the slope finder

- amistre64

the product rule for xy:
Dx(xy) = x'y + xy'

- amistre64

x' = 1 so its usually discarded

- amistre64

the power rule for y^5:
y' 5y^4

- anonymous

so my dy/dx on that would be 5? because it's changed for a total of 5 in the coeficent?

- amistre64

your dy/dx inthat one is undertermined until you kow what to plug in for a value of y

- amistre64

but when you do, then you can determine dy/dx = y'

- amistre64

5xy^2 = z derive with respect to t?

- amistre64

5x2y y' + 5 x'y^2 = z'

- amistre64

there are no "t" values so all your derivative bits stay put

- amistre64

y' = dy/dt in this case
x' = dx/dt
z' = dz/dt

- 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

the: 2sqrt(x) + sqrt(y) =5 one?

- anonymous

or\[2\sqrt{x}*1/2+1/2y^1/2\]

- amistre64

Dx(2sqrt(x)) + Dx(sqrt(y)) = Dx(5)

- amistre64

x' (1/sqrt(x)) + y' (1/2sqrt(y)) = 0

- amistre64

x' = dx/dx = 1 so we can ignore it.... or leave it in.
solve for y'

- amistre64

there is no f'(x)*f(x) + dy/dx...... your mixing notations and confusing yoursoef :)

- 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

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

- amistre64

solve for dy/dx which is another notation for y'

- anonymous

Alright i think i am getting it now. Fantasitic answer as always thanks a ton

- 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

Yea i think i was reading into it more than it was really asking me to

- amistre64

y^2 = 5x^2 + 6x +3

- amistre64

\[\frac{dy}{dx}2y = \frac{dx}{dx}10x + \frac{dx}{dx}6 +\frac{dx}{dx}0\]

- amistre64

y' 2y = 10x +6

- amistre64

10x + 6
y' = ---------
2y

- amistre64

its just deriving like normal; just keep the bit in that you were taught to throw out lol

- anonymous

OH!!! so it's the change in y over the change in x

- amistre64

yes :)

- anonymous

i see what you did there

- amistre64

no you see what youve always done there ;)

- 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

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

Well if they did things the easy way there would be a lot of math teachers out of a job.

- amistre64

lol maybe :)

- anonymous

Thanks for your time great explianition.

- 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

youre welcome :)

- 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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