00:01
Hi, for this problem we're supposed to find the d -y -d -x at 2 .0 and we're going to do that using implicit differentiation.
00:11
Now we have to effect on the derivative of each side and on the right -hand side we have four times we move four out and derivative e to the y is still e to the y but we need to multiply with d -y because changeful.
00:28
And for the second, for the right -hand side, inside we use product room so it's going to be the derivative of x square which is 2x multiplied by y to the fifth plus x squared is the same derivative y to the fifth is going to be 5 y to the fourth multiplied by dy -dx because of chain rule now we want to combine the term that have d -y -d -x together and we're going to do that by subtracting x square times y 5 y to the fourth d -y d -x from both sides so we're going to have four times e to the y d -y d -d -x plus minus of 5 x square y to the fourth d -y over d x equal to 2x to the 5th.
01:32
Now we're going to factor the d -ydx from the left -hand side.
01:36
So divide over the x multiply by 4e to the y minus 5x square, y to the 4th, decode to 2x times y to the 5th.
01:50
Now we're going to divide each side by 4 times e to the y, minus 5 times x square, y to the 5, to the fourth to isolate the y d x.
02:03
To the y of the x it's going to be equal to 2x times y to the fifth over 4 times a to the y minus 5x square y to the fourth.
02:17
So now that we know the x we're going to evaluate it at point 2 .0...
