00:01
High from the question given that consider the system of equation 2y1 minus y1 y2 cube plus y2 square plus x y1 is equal to 3 and y1 plus x y2 minus y1 cube y2 square plus 2x square will be equal to 3.
00:40
So here we need to find the dy1 by dx.
00:44
So here we need to use the implicit differentiation.
00:47
So first from the first equation 2 times of dy1 by dx minus y2 cube remains as a constant.
01:01
So here dy1 by dx plus 0 plus x times of dy1 by dx plus y1 of 1 will be equal to 0.
01:16
So here dy1 by dx will be multiplied with 2 minus y2 cube plus x will be equal to minus y1.
01:30
So here dy1 by dx is equal to minus y1 divided by 2 minus y2 cube plus x.
01:42
Now dy1 by dx at 1 comma 1 comma 1.
01:52
So this is nothing but y1 y2 x.
01:56
So here minus 1 divided by 2 minus 1 plus 1.
02:03
So this plus 1 minus 1 will get cancelled.
02:06
So we obtain dy1 by dx at 1 comma 1 comma 1 will be minus half.
02:16
Now let us move on to the second equation...