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Today in microeconomics class the following question was covered by the tutor:

If $z = x^2 + y^2$ changes from $(1,2)$ to $(1.05,2.1)$, compute the value of $\Delta z$ and $dz$.

If we write $f(x,y)=z$, then simply $\Delta z = f(1.05,2.1) - f(1,2)$.

However in the case of $dz$ things get interesting. The tutor derived the formula:

$$ df(x,y) = \frac{d}{dx}f(x,y)dx + \frac{d}{dy}f(x,y)dy \\ = 2xdx + 2ydy $$

Then we filled in $x=1,y=2,dx=0.05,dy=0.1$ to get $dz=df(1,2)=0.5$ (actually we arrived at $0.9$ but I can't see how that happened). The idea is that it is an approximation of $\Delta z$.

I was under the impression that you weren't allowed to use notation like this, to assign values to e.g. $dx$ . When I pressed the tutor about it he said he doesn't think it's mathematically sound but that this is what the lecturer wants to see.

Tim
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2 Answers2

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The notation is a little strange - I don't like "find the value of $dz$" - but the underlying mathematics is sound. It's distinguishing between the actual change in $z$ and a linear approximation to that change determined by the derivatives. I'll write it out for one variable (you seem OK with the partial derivatives needed for two).

The definition of the derivative is $$ f(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} $$ so for $h$ near $0$ you have the approximation $$ f(x+h) -f(x) \approx f'(x)h . $$ The left member is the actual change in value. The right member is the linear approximation - what you get by following the tangent line rather than the curve for a short distance. If you now rename $h$ as $dx$ to suggest that it's very small and rename $f(x+h) -f(x) $ as $\Delta f$ then you get a result like the one you quote.

This is a good way to reason in applications in physics and economics. It can be made rigorous when necessary.

Ethan Bolker
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  • You say you think the notation is strange. But by strange do you mean wrong/abusing notation? I get that it's supposed to be an approximation of the first order for the change of z. – Tim Nov 03 '22 at 15:02
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    I've never seen "calculate $dz$" to mean "calculate the linear approximation to $\Delta z$". It's not "wrong" in this context if the meaning is made clear. – Ethan Bolker Nov 03 '22 at 17:45
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What you've got there is the total differential of the function $f(x,y)$.

This is a non-rigorous notion in standard analysis, but can be made rigorous in nonstandard analysis.

For a function $f(\boldsymbol{x})=f(x_1, x_2, \dots, x_n)\,$, we define the total differential to be

$$\mathrm{d}f := \sum_{i=1}^n{\frac{\partial f}{\partial x_i}\,\mathrm{d}x_i}$$

This can be thought of nonrigorously as a linear approximation of $f$ about $\boldsymbol{x}=(x_1,x_2,\dots,x_n)$. A more standard notion that is mathematically equivalent is the Jacobian matrix (in one dimension, the gradient) of the function: we define

$$\nabla f := \left(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\dots,\frac{\partial f}{\partial x_n}\right)$$

Then, $(\nabla f)^\intercal$ is a linear map which is the best linear approximation to $f$ about $\boldsymbol{x}$ such that

$$(\nabla f)^\intercal \Delta\boldsymbol{x} = \sum_{i=1}^n{\frac{\partial f}{\partial x_i}\,\Delta x_i}$$

The total differential can be thought of as the infinitesimal version of this.

Lemmon
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