# Multivariable Linear Approximation – An expression with a power in 2 variables – Exercise 3390

Exercise

Find an approximate value of

$$1.02^{4.05}$$

$$1.08$$

Solution

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

$$x=1.02, y=4.05$$

because these values appear in the question. And we set

$$x_0=1, y_0=4$$

because they are the closest values to x and y respectively that are easy for calculations.

After we have defined x and y, it is easy to find the function. Just put x instead of the number we set to be x and put y instead of the number we set to be y. In this way, we get the function

$$f(x,y)=x^y$$

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

$$f'_x(x,y)=yx^{y-1}$$

$$f'_y(x,y)=x^y\ln x$$

We put all the data in the formula and get

$$f(1.02,4.05)\approx f(1,4)+f'_x(1,4)\cdot(1.02-1)+f'_y(1,4)\cdot(4.05-4)=$$

$$=1^4+4\cdot 1^3\cdot 0.02+1^4\ln 1\cdot 0.05=$$

$$=1+4\cdot 0.02+1\cdot 0\cdot 0.05=$$

$$=1+0.08+0=$$

$$=1.08$$

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