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

11+xy1x+y\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)f(x0,y0)+fx(x0,y0)(xx0)+fy(x0,y0)(yy0)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,x0,y0,f(x)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

x0,y0x_0,y_0

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

In our exercise, we will define

x0=0,y0=0x_0=0, y_0=0

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

f(x,y)=11+xyf(x,y)=\frac{1}{1+x-y}

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

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

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

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

We put all the data in the formula and get

11+xyf(0,0)+fx(0,0)(x0)+fy(0,0)(y0)=\frac{1}{1+x-y}\approx f(0,0)+f'_x(0,0)\cdot(x-0)+f'_y(0,0)\cdot(y-0)=

=11+00+1(1+00)2x+1(1+00)2y==\frac{1}{1+0-0}+\frac{-1}{{(1+0-0)}^2}\cdot x+\frac{1}{{(1+0-0)}^2}\cdot y=

=11x+1y==1-1\cdot x+1\cdot y=

=1x+y=1-x+y

Hence, we get

11+xy1x+y\frac{1}{1+x-y}\approx 1-x+y

As required.

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply