00:01
In this question, we are told that u of xy equals to e to the y times f capital over x times e to the negative y squared.
00:10
And in the first case, for f capital equals to l n, we are asked to calculate d u over d x and d u over d y.
00:19
So, we are told that f of z equals to ln z.
00:25
And in the first case, and in our case, z equals to x e to the negative y squared.
00:39
And now we are let's calculate therefore we can rewrite the function u of xy as e to the y multiplied by l n z right and this equals to e to the y times ln of x e to the negative y squared because that's what z is in the first case right because that's why z is.
01:12
Now let's calculate d u over d x.
01:17
D x equals to the derivative with respect to x of e to the y times ln z whereas z is a function which depends on x and y right here z contains both x and y and by the chain rule this equals so first of all e to the y is a constant with respect to x so we can factor it out we can rewrite this as e to the y times d over d x of lnz, where z is a function of x and y.
02:03
And by the chain rule, we can rewrite this as e to the y, multiplied by a d of lnz.
02:16
Sorry, i need to use straight d here.
02:19
D of lnz over a dz, multiplied by dz over dx.
02:29
That's by the chain rule, right? now the derivative of lnz is 1 over z.
02:39
Now let's calculate dz over dx.
02:42
Recall that z equals to x times e to the negative y squared.
02:47
Therefore, dz over dx equals to e to the negative y squared.
02:58
And remembering that z equals to x times e to the negative y squared, we are going to get that this equals to e to the y times e to the negative y squared, divided by x times e to the negative y squared.
03:13
So we replaced z by its expression in terms of x and y.
03:19
E to the negative y squared cancels, and we are going to get e to the y over x.
03:27
That's d u over d x.
03:30
Now let's calculate d u over dy.
03:35
D u over d y equals to d over d y of e to the y times l and z.
03:51
For convenience i'm going to write down z in the corner z equals to x times e to the negative y squared all right and this equals to now y is a main variable and x is a constant with respect to y and we are going to use the product rule here this equals to d over d y of e to the y times l n z plus e to the y multiplied by d over d y of l n z that's by the product rule next the derivative of e to the y with respect to y is e to the y itself times ln z plus e to the y now to calculate the derivative of ln z with respect to y we will use the chain rule this equals to d of ln z over d z multiplied by d z over d y y this equals to e to the y times lnz plus e to the y and the derivative of lnz with respect to z is 1 over z.
05:23
Now, n times dz or dy.
05:25
Now we need to calculate dz or dy to finish calculating d, d, d, u over d.
05:34
So dz over dy equals to the derivative with respect to y of the function z.
05:44
Which recall equals to x e to the negative y squared.
05:52
X is a constant with respect to y, so we can factor it out.
05:56
And then we need to differentiate e to the negative y squared.
06:03
And that's a one variable derivative, right? so this equals, this whole thing equals to x multiplied by negative 2y times e to negative y squared...