# Prove $\exists\theta\in(0,1)$ s.t. $\Delta f=\frac{\partial f}{\partial x}\Delta x+\frac{\partial f}{\partial y}\Delta y$

Let $$f(x,y)\in C^1$$ in $$\mathbb{R^2}$$ and let $$(x_0+\Delta x,y_0+\Delta y)$$ and $$(x_0,y_0)$$ be points in $$\mathbb{R^2}$$.

Prove that $$\exists\theta\in(0,1)$$ such that: $$f(x_0+\Delta x,y_0+\Delta y)-f(x_0,y_0)=\\\frac{\partial f}{\partial x}(x_0+\theta\Delta x,y_0+\theta\Delta y)\Delta x+\frac{\partial f}{\partial y}(x_0+\theta\Delta x,y_0+\theta\Delta y)\Delta y$$

At first glance, this seemed like the differentiability definition, but I tried to make the connection and unfortunately failed. I guess that MVT hides here, but I don't see how to rigorously reach it.

Thanks!

Note: I found a solution to this problem online (not here) but I couldn't understand it, so I'd really appreciate a somewhat detailed solution.

## 1 Answer

Use $$a$$ and $$b$$ for the endpoints, for convenience. Define $$g:[0,1]\to \mathbb R$$ by $$g(t)=f(bt+(1-t)a).$$ Then, by the mean value theorem and the chain rule, there is a $$\theta\in (0,1)$$ such that

$$g(1)-g(0)=g'(\theta)(1-0)=\nabla f(g(\theta)) \cdot (b-a)$$.

I'll let you unravel this to obtain your result.

• First of all, thank you very much! Now - I thought that I could say, that there exist $t_1,t_2\in(0,1)$ such that $x*=x_0+t_1 h$ and $y*=y_0+t_2 k$, but how do I show that $t_1=t_2\triangleq\theta$? Also, even if I manage to prove that, I am not exactly left with that I need to prove – Amit Zach Apr 18 at 22:14
• I misread your question sorry! I will repost. – Matematleta Apr 18 at 22:22
• Now I understand. Thank you very much! – Amit Zach Apr 19 at 7:52
• You´re welcome! – Matematleta Apr 19 at 13:50