00:06This video develops some exercises on second-order partial derivatives.
00:11This also includes an explanation of the Jacobian matrix, which is also part of
00:16higher-order derivatives or partial differentials, due to their use in double integrals,
00:21triples. The Jacobian matrix is the application of first-order partial derivatives, in which
00:27The elements are arranged thus forming a determinant, ordered in rows and columns.
00:32In space, it can be scalar or vector, by assigning rectangular values in a given
00:37point in x, y, z, once the matrix is developed. For this we have the following exercise, if the
00:44function f equals x, plus 3y squared, minus z squared, function g equals 2x
00:51squared, y, z, the function h is equal to 2z squared, minus, x, y, find the determinant
00:56Jacobian of the differential of the functions fgh, on x, y, z, at the point 1, minus 1, 0.
01:04To do this, we determine each element in rows and columns, and designate each function f1, f2,
01:10f3, so for the first element, we take the partial derivative of f1 over the partial derivative of
01:15x when y and z are constants, this is the same differentiating for x we have 1, now we do the
01:21partial derivative of f1 over the partial derivative of y, when xz, are constants, differentiating
01:27For y, we have 6y. Similarly, we perform the partial derivative of f1 over the derivative
01:32partial of z when x and are constants, this by differentiating, we obtain minus 3z squared.
01:38Now, differentiating for the function f2, we have that the partial derivative of f2 over the derivative
01:43The partial derivative of X when YZ are constants is equal to the derivative of 4XYZ. In the function F2, the derivative
01:51partial derivative of F2 on the partial derivative of Y when XZ are constants, differentiating for Y we have 2X to
01:58square Z. The partial derivative of F2 over the partial derivative of Z when XZ are constants,
02:05Differentiating for Z, we obtain 2X squared Y. Differentiating for the function F3, we have that
02:15The partial derivative of F3 on the partial derivative of X when YZ is constant is equal to differentiating
02:21to Y. In the function F3, the partial derivative of F3 on the partial derivative of Y when
02:27XZ are constants, differentiating for Y we get X. The partial derivative of F3 on the derivative
02:35partial of Z when XY are constants, differentiating for Z, we obtain 4Z. Once obtained the
02:42values, we form the Jacobian determinant, for each value X, Y, Z. Solving by the
02:49Sarros' method, a 3x3 matrix, we have the first positive diagonal, we have 0, the second
02:55diagonal we have 0, the third diagonal we have 12, for the negative diagonals, in the first
03:00We have negative 0, the second negative diagonal; we have negative 0, the third negative diagonal.
03:06We have minus 2, we get 10.
03:13Second-order partial derivatives describe the rate at which the derivative itself changes.
03:17partial with respect to its variables. A function F of two independent variables
03:22X, Y, has two first-order partial derivatives, F, X, F, Y, each of these derivatives
03:29First-order partial derivatives have two partial derivatives, giving a total of 4 second-order partial derivatives.
03:34partial derivatives.
03:36Now we have the following exercise given the function, Z equals X cosine of Y, minus y
03:41cosine of X, find all the second-order partial derivatives, for this we recall the derivative
03:47of the product of two functions, which is equal to, U times the derivative of B, plus, B times the derivative
03:52From U, as well as the rules for differentiating sine and cosine, we have the partial derivative
03:57of Z on X, when Y is constant, using the derivative of the product and differentiating for X,
04:03We have cosine Y, plus Y sine of X, and also the partial derivative of Z with respect to the derivative
04:09partial of Y, when X is constant, is equal to and differentiating by product differentiation,
04:14We have, minus X sine of Y minus cosine of X.
04:19Now we have the partial derivative of Z on X, this is equal to the derivative of the product
04:24of two functions, and is equal to Y cosine of X.
04:29Now we have the second partial derivative of Z with respect to X, when Y is constant, that is
04:34equal to the derivative of the product of two functions, and is equal to Y cosine of X.
04:40We also have the second partial derivative of Z over the product of the differentials
04:45partials of YX, when X is constant, this by differentiating by the product and using the rules
04:51From sine and cosine we obtain, less sine of Y, plus sine of X.
04:55Continuing with the exercise, the second partial derivative of Z with respect to Y, when X is constant,
05:01It is equal to the derivative of the product and the sine and cosine rules, we have as our answer
05:05minus X cosine of Y, differentiating the second partial derivative of Z over the product of
05:10the differentials of XY, and by differentiating we obtain the answer, less sine, more sine of X,
05:15and the equality in partial derivatives is observed.
05:19That's all, if you found the video helpful, don't forget to subscribe.
05:23Thank you!
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