Suppose f(x, y) = origin. fx(0,0) = fy(0,0) = { sin(8x³+2y4) 5x²+7y² 0, 2 (Enter "DNE" if the derivative does not exist.) (x, y) = (0,0) (x, y) = (0, 0). Find the partial derivatives of f at the

Find the value of at the point (1, 3, 1) if the equation 6xz + ylnx – 5x2 –1 = 0 defines x as a function of the two independent variables y and z and the partial derivative exists. (x,y,z) = (1,3,1)

Let – 4x3 – 2y³ – 5z3 – 4xyz + 619 = 0. dz dz and at the point (6, 5, – 3). - Use partial derivatives to calculate ду az (6,5,-3) az ду (6,5,-3)

Find all the second order partial derivatives for f(r, y) = sin(2x) – a²e" + y?. Verify that the conclusion of Clairut's Theorem holds, that is fry = fyz.