Let us form a Lagrangian of this problem. Suppose that the conditions under which this problem cannot be solved hold. That means that all functions that are in the problem are continuously differentiable, that the solution of this problem exists. All Lambda's also exists and in this condition also holds. Then when we are asked, how to find derivative or the value function? The value function for this particular problem will be the profit function. So, after solving the problem, we need to substitute into the profit. The found values, and since they depend only on the resources on b's. So, that will be a function, a profit function, pie star, which depends only on b's. Now, if we're interested in finding derivative with respect to any, let it be jth resource. According to the envelope theorem, we differentiate Lagrangian with respect to this constraint, the value of b, and we get Lambda j star. So, it tells us the value how important for the production is this particular resource. That's why Lagrangian multiplier, Lambda jth, is called shadow price of jth resource. Now, if we recall the very first envelope theorem, which concerned the unconstrained optimization, we had to make an additional assumption we would like to relax right now. What I am talking about is, given an objective function depending on x and Alpha, which we need to maximize with respect to x. When we solve it, we get the maximizer which is a function of Alpha, and we assumed that for each Alpha from some interval on the number line, these n functions, because this is a vector from the n-dimensional space, this vector function is continuously differentiable with respect to Alpha. This is an additional condition which we can replace by something different. Okay. I'll show you that if we, suppose from the start, at our function is twice continuously differentiable. There is no big limitation here, because the majority of functions we use in macro and microeconomics are quite smooth. Let's suppose that the second order condition in the form of the Hessian function is also fulfilled is valid in the form of the definiteness of the Hessian matrix. So, let us recall that given the maximizer, or rather this is a critical point of the objective function, f, we state that the Hessian matrix of f with respect to x is negative definite. We can state it shortly with this inequality, less than 0. Now, I will show you that under these conditions, the first condition being that x star is a solution of the first-order conditions, which is a system of derivatives of the objective function with respect to all x's. This derivative should be 0, and also additionally, the second order condition in the form of the negatively definite Hessian also holds. Then, the critical point x star as a function of Alpha isn't necessarily a continuously differential function. How to see that? We need to write down the first-order conditions, and we may consider all these n equations as a system of equations defining M implicit functions. So, we need to apply IFT to all these equations or rather to the system of equations. This IFT theorem, which were regarded as the most complicated in terms of stating the results, has a condition under which this system is solvable, and the solution will be a function of parameter Alpha and this function or these n functions are continuous differentiable. In terms of differentiability, all these derivatives are continuous differentiable because the original function belongs to c2 class of functions. But we need to check that the determinant of the Jacobian matrix based on this system of equations has a non-zero determinant at a particular point. So, how this Jacobian matrix will look like? So, J, Jacobian matrix, this time is a square matrix because we have n variables and we have n equations here. So, we take the first equation and we differentiate with respect to all x's starting with the first. That's how would get f_11. After that we differentiate with respect to x_2, that becomes 21 and we reach F_n1 since Young's theorem is applicable because f belongs to C_2. This is not exactly the Hessian matrix or the function f or rather a transpose of this matrix. But it doesn't affect the determinant of J of course. So, you proceed filling in the entries. The final will be when we take the final equation and start differentiating with respect to all x's. So here we have 1n, 2n, until f_nn. So, this matrix being the Jacobian matrix based on this system of equations. At the same time is the Hessian matrix of the function considered here. Since we assumed initially that the second order condition is valid, that means the Hessian metrics of F is negative definite. Then, the biggest leading principle minor, all of these matrix should be a non-zero the number. The sign doesn't matter for us, but we know the sign is the same as minus 1 raised to the nth power. So, clearly, determinant of J is not 0. Then, it implies that x star is a function of Alpha not only exists about locally in the neighborhood of a point, but also is continuous differentiable.