00:01
Okay, so i'm not going to write the whole question because we don't need all of the information in there.
00:06
But essentially we're told that, so suppose we have a big function f and we want to consider an equation of the form f of xy equals c.
00:14
And we're told that this implicitly defines y and the function of x, right? at the neighborhood of a point, it doesn't matter for the first question.
00:26
And we're asked to prove then using the chain rule that if f is differential, then so is y, this function implicitly defined, and that the derivative, the y, the x, can be obtained from the derivatives of f as minus derivative of x at the point xy over the derivative of y at the point x, at the point x, y, at the point x, y, over the derivative of f with respect to y, xy, and this is an application of the chain rule, right? so if, so we are working at the neighborhood of a point, where y is defined as a function of x, then what we know is that at that neighborhood, within that neighborhood of that point, we have f of x, y of x, is equal to zero.
01:12
So on the left hand side, we have a function of x, right? so differentiating with respect to x, what we get is, so we need the derivative of f with respect to the first component, vf the x at the point xy, let me not write xy of x.
01:31
Of the first component times the derivative of the first component with respect to x this will just be 1 plus derivative of f with respect to the second component times the derivative of the second component with respect to x right and this is equal to 0 and now we know how these things show up right of course this is here is equal to 1 and here this is simply the y the x because we are assuming that we can write y as a function of x so this simplifies without writing the point now as the fdx plus the fd y times the y the x is equal to zero so if f is a differential all of this makes sense and in particular now if d fd y is not zero this is the same as then saying that d y the x is so minus the fd x goes to the other side the fd x over the fd y if the y is not zero all right so these proof of you wanted to prove so at the neighborhood of the point where we can write y as a function of x if f is differentiable and if the fdy is not zero.
02:37
And indeed, this statement here is true.
02:41
And now part b, you can look at the picture on the question sheet, but part b essentially says, now let's take f of x y as 2x cubed y minus y to the fifth x, right? and we're claiming that f of xy equals 1 defines y as a function of x around the point uh one one and we're asked to find the y the x okay so this is essentially what the person says and you can see a nice picture to understand why you really need to read locally so the function there's no function y of x defined globally but around the point one one there is and let's just really use the formula right so what we just saw was that the y d x at 1 will be minus the fd x at 1 .1 over the f the y at 1...