We're going to compute the local linear approximation of F f P and use that to approximate the value of FX que And then from there we will take the absolute value of our local linear approximation. Evaluated a Q minus f evaluated at Kew and divide that by the distance between P and Q. So to get started, let's find the partial derivative of F with respect to acts on the partial derivative of F with respect toe wine. So we know that f of X Y is equal to the quantity X squared plus y squared to the minus one half. So the partial derivative of F with respect to X is equal to minus one half times of quantity. X squared plus y squared to the minus three halfs times two acts. And now that we have this, we can evaluate the partial derivative of F with respect to X. At the point, P and P is the 0.4 comma three. And when we do that, we get that this is equal to minus four all over 125 and similarly, when we compute the partial derivative of F with respect to why we get that This is minus one half times a quantity X squared plus y squared to the minus three halfs this time multiplied by two y. And when we evaluate this at P, we get that this is equal to minus three over 125. So now from here, we know that our local linear approximation of F at the point p that's equal to F evaluated at P plus the partial derivative of F with respect to X evaluated at p times a quantity x minus x not which in this case is for plus the partial derivative of f with respect to why evaluated at P times of quantity, why minus why not, which in this case is three but we solved for these values above. So all we have to do is just plug in that information we get. That local linear approximation is equal to 1/5 minus 4/1 25 times the quantity X minus four minus 3/1 25 times the quantity. Why minus three. So that completes part one. And now for part two, we just plug in. Q. We plug Q into our local linear approximation and into F and take that difference and then dividing by the distance between P and Q. So when we do that, we get that. That's the absolute value of 1/5 minus 4/1 25. And then in this time, instead of writing X well, right? The X coordinate of Q, which is 3.92 And then this time instead of writing why? Well, right, the Y coordinate of Q, which is 3.1 And then from here we want to subtract F evaluated at Kew. So that's just one over the square root of 3.92 squared, plus 3.1 squared. And then we want to divide that by the distance between P and Q. Which is the square? The sum of the square of these numbers above. So we get that. That's 3.92 minus form squared, plus the quantity 3.1 minus three squared. And then once we compute the value of this, we get that that's approximately 0.0 17 66 036 and that completes the problem. But more importantly, our reason for computing. This value was to show that the air in our approximation is much smaller than the distance between are two points p and Q. Thanks for watching.