Given the function of two variables z yexy then the second...

Given the function of two variables, $z = ye^{xy}$ Then the second derivative, $f_{xy}$, is equal to: $ye^{xy}(2 + xy)$ $xy^3e^{xy}$ $xe^{xy}(2 + xy)$ $ye^{xy}(1 + xy)$ $xe^{xy}(1 + xy)$ Clear my selection

Transcript

00:01 Hi, in this question we have been given a function f of xy is equal to x y squared plus y into t raised to the power x squared plus five and we need to find all the partial derivative of this function.

00:20 So let us find the derivative of f with respect to x.

00:25 So since we are finding partial derivative with respect to x, so here we will keep y constant.

00:30 So this will become y squared into 1 plus again here we will give y constant and we will differentiate with respect to x so this will become e -res to the power x squared into 2x and the derivative of constant is 0.

00:46 So here we are using this concept that is derivative of e -r -rish -to -the -power x is always equal to e -r -r -to -the -power x and derivative of x -r -rease to the power n is equal to n times x to the power m minus 1 so from here this partial derivative with respect to x would be y squared plus so this will become 2 x y e range to the power x is y again we will differentiate with respect to x only so this will become del 2 f of del x2 is equal to so again we will keep y constant so this if this is a constant here so the derivative will become 0 and here we will take y outside and also this 2 outside and we will have x into e -raged to the power x squared so first of all we will keep e -raged to the power x squared and we'll differentiate x so this will be 1 plus now we will keep x constant and differentiate e -rage to the power x squared so this will become e -r -r -r -r -to -the -power x squared into 2x so if we arrange this we will get double derivative of f with respect to x is equal to so here we can take e -dh to the power x is y so this will become 2 y e -re h -rish to the power x squared 1 plus 2 x so we will consider this as equation one.

02:00 Now we will find derivative of f with respect to y.

02:04 So this will be now here we will keep x constant.

02:07 So this will become x into 2y and here this will be e.

02:12 To the power x squared into 1 plus the derivative of constant is 0.

02:17 So we will get partial derivative of f with respect to y is equal to 2xy plus e to the power x squared.

02:26 Again we need to find the double derivative.

02:28 So the double derivative with respect to y.

02:31 So again, we will keep x constant...