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If $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is differentiable at the point $(x_0,y_0)$ then $$ \lim_{t\to 0} \frac{f(x_0+tx,y_0+ty)-f(x_0,y_0)}{t}=xf_x(x_0,y_0) + yf_y(x_0,y_0) $$

I know if the function is differentiable then the partial derivatives exist so we could use $$ \lim_{t\to 0} \frac{f(x_0+t,y_0)-f(x_0,y_0)}{t} = f_x(x_0,y_0) $$ and $$ \lim_{t\to 0} \frac{f(x_0,y_0+t)-f(x_0,y_0)}{t} = f_y(x_0,y_0) $$ but I can't figure out how to deal the $f(x_0+tx,f_0+ty)$ term. Am I approaching the proof incorrectly? How could I prove this?

Nathan
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1 Answers1

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Let $u(t)=x_0+tx$, $v(t)=y_0+ty$, and $g(t)=f(u(t),v(t))$.

$$\lim_{t\to 0} \frac{f(x_0+tx,y_0+ty)-f(x_0,y_0)}{t} = \lim_{t\to 0} \frac{g(t)-g(0)}{t}=:g'(0)$$

But, by Chain Rule (roughly re-justified below), we have that: $$ g'(t) = \lim_{h\to 0}\frac{g(t+h)-g(t)}{h} $$ $$ = \lim_{h\to 0}\frac{f(u(t+h),v(t+h))-f(u(t),v(t+h))}{h} + \lim_{h\to 0}\frac{f(u(t),v(t+h))-f(u(t),v(t))}{h} $$ $$ = \lim_{h\to 0}f_x(u(\xi_u),v(t+h))u'(\xi_u) + \lim_{h\to 0}f_y(u(t),v(\xi_v))v'(\xi_v) $$ $$ =f_x(u(t),v(t))u'(t) + f_y(u(t),v(t))v'(t), $$ with $\xi_u$ and $\xi_v$ between $t$ and $t+h$ given by Mean Value Theorem.

Setting $t=0$, we get the result (directional derivative).

ir7
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