 Posers and Puzzles emanon Posers and Puzzles 21 Apr '07 13:18
1. 21 Apr '07 13:18
I'm having problems with this exercise on implicit differentiation:

If K^1/3 L^1/3 = 24, compute dL/dK by implicit differentiation.

What I get so far:

1/3K^-2/3 L^1/3 + 1/3K^1/3 L^-2/3 (dL/dK) = 0

Then solving for (dL/dK) = (1/3K^-2/3 L^1/3) / -(1/3K^1/3 L^-2/3)

But I don't know how to solve the last part, (1/3K^-2/3 L^1/3) / -(1/3K^1/3 L^-2/3) that is....🙁
2. 21 Apr '07 16:12
Originally posted by emanon
I'm having problems with this exercise on implicit differentiation:

If K^1/3 L^1/3 = 24, compute dL/dK by implicit differentiation.

What I get so far:

1/3K^-2/3 L^1/3 + 1/3K^1/3 L^-2/3 (dL/dK) = 0

Then solving for (dL/dK) = (1/3K^-2/3 L^1/3) / -(1/3K^1/3 L^-2/3)

But I don't know how to solve the last part, (1/3K^-2/3 L^1/3) / -(1/3K^1/3 L^-2/3) that is....🙁
1/3K^1/3 L^-2/3 (dL/dK) = -1/3K^-2/3 L^1/3 +

dL/dK=-L/K

The 1/3 simplifies and the powers add up to 1 and -1 (for L and K, respectively).

Economics student?
3. 21 Apr '07 16:33
Originally posted by Palynka
1/3K^1/3 L^-2/3 (dL/dK) = -1/3K^-2/3 L^1/3 +

dL/dK=-L/K

The 1/3 simplifies and the powers add up to 1 and -1 (for L and K, respectively).

Economics student?
How did you get the ...+ dL/dK did it just magically appear there or....

Neither do I understand how the powers add up to -1 and 1, could you possibly explain that?

yeah, economics....*cry*
4. 21 Apr '07 16:58
Originally posted by emanon
How did you get the ...+ dL/dK did it just magically appear there or....

Neither do I understand how the powers add up to -1 and 1, could you possibly explain that?

yeah, economics....*cry*
K^1/3 L^1/3 = 24

Through differentiation you get:

1/3*K^(-2/3)*L^(1/3)dK+1/3*L^(-2/3)*K^(1/3)dL=0

pass the first element to the RHS, divide both sides by 1/3

L^(-2/3)*K^(1/3)dL = -K^(-2/3)*L^(1/3)dK

multiply both sides by L^(2/3)*K^(-1/3)

dL = -K^(-2/3)*k^(-1/3)*L^(1/3)*L^(2/3)dK

divide both sides by dK and factor K and L

dL/dK = -K^(-2/3-1/3)*L^(1/3+2/3)

dL/dK = -L/K
5. 21 Apr '07 17:02
Originally posted by Palynka
K^1/3 L^1/3 = 24

Through differentiation you get:

1/3*K^(-2/3)*L^(1/3)dK+1/3*L^(-2/3)*K^(1/3)dL=0

pass the first element to the RHS, divide both sides by 1/3

L^(-2/3)*K^(1/3)dL = -K^(-2/3)*L^(1/3)dK

multiply both sides by L^(2/3)*K^(-1/3)

dL = -K^(-2/3)*k^(-1/3)*L^(1/3)*L^(2/3)dK

divide both sides by dK and factor K and L

dL/dK = -K^(-2/3-1/3)*L^(1/3+2/3)

dL/dK = -L/K
🙄😲
6. 21 Apr '07 17:10
Originally posted by Palynka
K^1/3 L^1/3 = 24

Through differentiation you get:

1/3*K^(-2/3)*L^(1/3)dK+1/3*L^(-2/3)*K^(1/3)dL=0

pass the first element to the RHS, divide both sides by 1/3

L^(-2/3)*K^(1/3)dL = -K^(-2/3)*L^(1/3)dK

multiply both sides by L^(2/3)*K^(-1/3)

dL = -K^(-2/3)*k^(-1/3)*L^(1/3)*L^(2/3)dK

divide both sides by dK and factor K and L

dL/dK = -K^(-2/3-1/3)*L^(1/3+2/3)

dL/dK = -L/K
😀 thx I got it now
7. 21 Apr '07 19:39
Originally posted by emanon
I'm having problems with this exercise on implicit differentiation:

If K^1/3 L^1/3 = 24, compute dL/dK by implicit differentiation.

What I get so far:

1/3K^-2/3 L^1/3 + 1/3K^1/3 L^-2/3 (dL/dK) = 0

Then solving for (dL/dK) = (1/3K^-2/3 L^1/3) / -(1/3K^1/3 L^-2/3)

But I don't know how to solve the last part, (1/3K^-2/3 L^1/3) / -(1/3K^1/3 L^-2/3) that is....🙁
Isn't is more simple just to cube everything and then work from there?

Don't you have KL=24^3 so

dL / dK = -24^3/K^2

?
8. 21 Apr '07 20:34
Originally posted by SPMars
Isn't is more simple just to cube everything and then work from there?

Don't you have KL=24^3 so

dL / dK = -24^3/K^2

?
That is the easier way to do it, but d(24^3)/dK = 0, so you end up with dL/dK = -L/K as before.
9. 21 Apr '07 20:49
Originally posted by SPMars
Isn't is more simple just to cube everything and then work from there?

Don't you have KL=24^3 so

dL / dK = -24^3/K^2

?
I don't get that at all...🙁
10. 21 Apr '07 23:49
His point is that you can actually solve for L in this one and find the derivative as an explicit function, involving only K. But cubing both sides is also an easier way to get the implicit derivative:

KL = 24^3
d/dK (KL) = 0
L + K(dL/dK) = 0
K(dL/dK) = -L
dL/dK = -L/K

If you plug in L = (24^3)/K from the first line, you get the answer SPMars gave.
11. 22 Apr '07 13:58
Originally posted by CZeke
His point is that you can actually solve for L in this one and find the derivative as an explicit function, involving only K. But cubing both sides is also an easier way to get the implicit derivative:

KL = 24^3
d/dK (KL) = 0
L + K(dL/dK) = 0
K(dL/dK) = -L
dL/dK = -L/K

If you plug in L = (24^3)/K from the first line, you get the answer SPMars gave.
For us not so brilliant in math, what is the differance between implicit and explicit functions?
12. 22 Apr '07 14:19
Originally posted by sonhouse
For us not so brilliant in math, what is the differance between implicit and explicit functions?
A function is a function, whatever the situation.

Whether it's 'explicit' or 'implicit' is to do with how it's written down.

F(x) = 2x+9

G(x)^5 + G(x) + x = 0

Both F and G are functions, but F is in 'explicit' form because we have an explicit formula for its values. G is in 'implicit' form because for G(x) we do not have such an expression.

Of course the difference between the two notions is purely cosmetic and if you're cunning enough an implicit function can be written explicitly (although the expression we get might be rather messy). With the G example above it's possible to express G in terms of its inverse function, which is easily found.
13. 22 Apr '07 14:20
Originally posted by SPMars
A function is a function, whatever the situation.

Whether it's 'explicit' or 'implicit' is to do with how it's written down.

F(x) = 2x+9

G(x)^5 + G(x) + x = 0

Both F and G are functions, but F is in 'explicit' form because we have an explicit formula for its values. G is in 'implicit' form because for G(x) we do not have such an expression.

...[text shortened]... it's possible to express G in terms of its inverse function, which is easily found.
14. 22 Apr '07 14:275 edits
Originally posted by sonhouse
Define a function H mapping R to R by

H(y) := -y^5-y

Since this is a strictly decreasing, continuous, bijection R -> R it has an inverse function G which is strictly decreasing and continuous and satisfies the relation G(x)^5+G(x)+x=0.

Furthermore, given x in R we have

G(x) = sup ( y: H(y) > x ).

But I don't expect you to understand the notation or the ideas I am using.
15. 22 Apr '07 14:39
Originally posted by SPMars
Define a function H mapping R to R by

H(y) := -y^5-y

Since this is a strictly decreasing, continuous, bijection R -> R it has an inverse function G which is strictly decreasing and continuous and satisfies the relation G(x)^5+G(x)+x=0.

Furthermore, given x in R we have

G(x) = sup ( y: H(y) > x ).

But I don't expect you to understand the notation or the ideas I am using.
Ok, what is 'bijection' and 'sup'? Terms I never heard of.