Exercise
Find an approximate value of
1.02^{4.05}
Final Answer
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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