21 Apr 07
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....🙁
21 Apr 07
Originally posted by emanon1/3K^1/3 L^-2/3 (dL/dK) = -1/3K^-2/3 L^1/3 +
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....🙁
dL/dK=-L/K
The 1/3 simplifies and the powers add up to 1 and -1 (for L and K, respectively).
Economics student?
21 Apr 07
Originally posted by PalynkaHow did you get the ...+ dL/dK did it just magically appear there or....
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?
Neither do I understand how the powers add up to -1 and 1, could you possibly explain that?
yeah, economics....*cry*
21 Apr 07
Originally posted by emanonK^1/3 L^1/3 = 24
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*
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
21 Apr 07
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
21 Apr 07
Originally posted by Palynka😀 thx I got it now
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
21 Apr 07
Originally posted by emanonIsn't is more simple just to cube everything and then work from there?
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....🙁
Don't you have KL=24^3 so
dL / dK = -24^3/K^2
?
21 Apr 07
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.
22 Apr 07
Originally posted by CZekeFor us not so brilliant in math, what is the differance between implicit and explicit functions?
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.
22 Apr 07
Originally posted by sonhouseA function is a function, whatever the situation.
For us not so brilliant in math, what is the differance between implicit and explicit functions?
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.
22 Apr 07
Originally posted by SPMarsSo easily find it please.
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.
22 Apr 07
5 edits
Originally posted by sonhouseDefine a function H mapping R to R by
So easily find it please.
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.
22 Apr 07
Originally posted by SPMarsOk, what is 'bijection' and 'sup'? Terms I never heard of.
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.