# Multivariable Linear Approximation – An expression with a fraction in 2 variables – Exercise 3402

Exercise

Prove that for x and y that are close to zero the following holds

$$\frac{1}{1+x-y}\approx 1-x+y$$

Proof

Since the equation we need to prove has the estimator sign, we probably need to use the linear approximation formula. Since there are two variables, we will use the 2-variable linear approximation formula

$$f(x,y)\approx f(x_0,y_0)+f'_x(x_0,y_0)\cdot(x-x_0)+f'_y(x_0,y_0)\cdot(y-y_0)$$

Therefore, we need to define the following

$$x, y, x_0, y_0, f(x)$$

And place them in the formula. x and y will be the numbers that appear in the question, and the points

$$x_0,y_0$$

will be the closest values to x and y respectively, which are easy for calculations.

In our exercise, we will define

$$x_0=0, y_0=0$$

The function will be the right side in the equation that we need to prove

$$f(x,y)=\frac{1}{1+x-y}$$

In the formula above we see the function partial derivatives. Hence, we calculate them.

$$f'_x(x,y)=\frac{-1}{{(1+x-y)}^2}$$

$$f'_y(x,y)=\frac{-1}{{(1+x-y)}^2}\cdot (-1)=$$

$$=\frac{1}{{(1+x-y)}^2}$$

We put all the data in the formula and get

$$\frac{1}{1+x-y}\approx f(0,0)+f'_x(0,0)\cdot(x-0)+f'_y(0,0)\cdot(y-0)=$$

$$=\frac{1}{1+0-0}+\frac{-1}{{(1+0-0)}^2}\cdot x+\frac{1}{{(1+0-0)}^2}\cdot y=$$

$$=1-1\cdot x+1\cdot y=$$

$$=1-x+y$$

Hence, we get

$$\frac{1}{1+x-y}\approx 1-x+y$$

As required.

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