Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

now dx = 1 and dy=y' solve for y'

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

Answer 8

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

y ' = 2sqrt(y)/sqrt(x)

Answer 9

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

Answer 10

i lost the negative lol

Answer 11

OK trying to process this really quick

Answer 12

its confusing at first becasue we are used to discarding that dx.... keep all the bits in and adjust for them in the end

Answer 13

y = 2x

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

Answer 14

dy is just another term for y'

Answer 15

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

Answer 16

sorta, cant say im familiar with your notation there

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

6y^1

Answer 21

you threw out the dy...how come?

Answer 22

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

Answer 23

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

Answer 24

no it doesn't

Answer 25

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

Answer 26

the y with respect to x doesnt disappear does it?

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

ok i'm with you

Answer 31

watch this one and Im sure your gonna slap yourself:

d (5x^3)
--------- = ?
dx

Answer 32

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

Answer 33

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

Answer 34

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?

Answer 35

dx/dx = 1

Answer 36

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

Answer 37

ya with me?

Answer 38

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

Answer 39

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

Answer 40

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 :)

Answer 41

not point slope but the slope finder

Answer 42

the product rule for xy:

Dx(xy) = x'y + xy'

Answer 43

x' = 1 so its usually discarded

Answer 44

the power rule for y^5:

y' 5y^4

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

y' = dy/dt in this case

x' = dx/dt

z' = dz/dt

Answer 52

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)

Answer 53

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

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

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

solve for y'

Answer 58

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

Answer 59

ok so it looks like you are just adding the dirivites inorder to solve this. am i hitting left field with that one?

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

y' 2y = 10x +6

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

yes :)

Answer 72

i see what you did there

Answer 73

no you see what youve always done there ;)

Answer 74

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

Answer 75

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.

Answer 76

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

Answer 77

lol maybe :)

Answer 78

Thanks for your time great explianition.

Answer 79

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

Answer 80

youre welcome :)

Answer 81

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.