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If I am given a function $f$ of two variables, where partial derivatives exist on all of $\mathbb{R}^2$. Can I apply the mean value theorem on an interval $[x_1,x_2]$ to $f(x,y)$ by fixing a $y_0$ and then apply the mean value theorem for functions of one variable to choose a $c \in (x_1,x_2)$ with $f(x_2,y_0)-f(x_1,y_0)=f'(c)(x_2-x_1)$?

Scott Frazier
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    Yes, for a fixed $y_0$, $f(x,y_0)$ can now be thought of as a function of just $x$. However, your $f'(c)$ doesn't quite make sense, since $f$ is a function of two variables. The line would be $f(x_2,y_0)-f(x_1,y_0)=f_x(c,y_0)(x_2-x_1)$. – ndhanson3 Mar 17 '21 at 19:36
  • @ndhanson3 So the last line you mention would be the correct statement? I am just a little confused with your phrasing – Scott Frazier Mar 17 '21 at 19:37
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    Yes, the equation I gave is the correct version of the equation you wrote. – ndhanson3 Mar 17 '21 at 19:39

3 Answers3

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In general for a differentiable function from $\mathbb{R}^2$ to $\mathbb{R}$, there exists a point $(c,d)$ on the line segment from $(x_1,y_1)$ to $(x_2,y_2)$ such that

$$f(x_2,y_2)-f(x_1,y_1)=\frac{\partial f}{\partial x}(c,d)(x_2-x_1)+ \frac{\partial f}{\partial y}(c,d)(y_2-y_1)$$

Taking $y_2 = y_1 = y_0$ the above expression simplifies to

$$f(x_2,y_0)-f(x_1,y_0)=\frac{\partial f}{\partial x}(c,y_0)(x_2-x_1)$$

RRL
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  • I can understand this intuitively, but how do I prove this formally? – Ambica Govind Feb 08 '22 at 13:22
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    @AmbicaGovind: You can find a proof of the mean value theorem for a multivariate function in standard real analysis textbooks. To see how such a proof works, introduce the function$\phi$ from $\mathbb{R}^2$ to$ \mathbb{R}$ given by $\phi(t) = (1-t)(x_1,y_1) + t (x_2,y_2)$ and define the composition $g(t) = f(\phi(t))$. Now apply the univariate mean value theorem to $g$ and get $f(x_2,y_2) - f(x_1,y_1) = g(1) - g(0) = g'(\xi)(1-0) = g'(\xi)$. Then use the chain rule to evaluate $g'(\xi)$ and you will get what I wrote in the first equation above. – RRL Feb 08 '22 at 17:34
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Yes, for a fixed $y_0$, $f(x,y_0)$ can now be thought of as a function of just $x$. However, your $f′(c)$ doesn't quite make sense, since $f$ is a function of two variables. The line would be $$f(x_2,y_0)−f(x_1,y_0)=f_x(c,y_0)(x_2−x_1).$$

ndhanson3
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Yes, but note that we are in 2 dimensions, so:

$$ f(x_2,y_0) - f(x_1,y_0) = \left. \frac{\partial f(x,y)}{\partial x} \right|_{(x,y)=(c,y_0)} (x_2-x_1) $$

nachosemu
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