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Verify the linear approximation at $ (0, 0) $.$ \dfrac{y - 1}{x + 1} \approx x + y - 1 $

$$L(x, y)=-1+x+y$$

Calculus 3

Chapter 14

Partial Derivatives

Section 4

Tangent Planes and Linear Approximations

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for this problem. Let's define the function to be the function on the left hand side. Why minus one over X plus one. And we want to verify the given linear approximation at the origin. So really, what we can we can reword This is we want to show that the linear ization of this function f at the point zero zero is given by this linear function over here. Subtle linear ization of the function X y. So this is a function of two variables here, just like f. Now the formula is in this case, we're using a equals B equals zero. That's the coordinates that are given. So you plug both of those into F and you also plug it into the partial derivative multiplied by X minus a and then also for the other partial derivative. With respect to why and then why minus b so that's the general formula being used with a equals B equals zero. So now f of zero zero, we can just plug that in into our function there, given by this formula here now, here will see we need the partial derivative of F with respect attacks. So first in this case, if you want, you could write the function not necessary to do this. But it might be easier to just use the power rule than teach the product. The question Well, here since X is in the denominator. So here, the partial derivative square that denominator. And when we do this, we see that the partial derivative of F with respect to X at the origin. So go ahead and plug in. Why equal zero? You got a one up top and you also get a one on the bottom. So that's just one. And then X minus zero is just X and similarly will need the partial derivative of effort there. Spencer, why, in this case, wise in the numerator. So the derivative of that it's just one, and we see f y zero zero is also one. So we're gonna have a plus one, and then why minus zero was just why and simplifying this term right here that were circling. We have negative one plus x plus y, and that's exactly the same as the linear function that was given in the beginning. So showing that the linear ization at the origin was X plus y minus, one verifies that given approximation. So that's our final answer

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