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$$z = f \left( u(t), \ v(t)\right) $$

In my lecture notes, it is said that

$$ \lim_{h\rightarrow0}\frac{f\left(u\left(t+h\right){,}\ v\left(t+h\right)\right)-f\left(u\left(t\right){,}\ v\left(t+h\right)\right)}{h}=\frac{\partial f}{\partial u}\frac{du}{dt}.$$

However I would have defined that derivative as the following limit:

$$ \lim_{h\rightarrow0}\frac{f\left(u\left(t+h\right){,}\ v\left (t\right)\right)-f\left(u\left(t\right){,}\ v\left(t\right)\right)}{h} $$

Could someone clarify how these limits are the same?

mathslover
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1 Answers1

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I think this is one of those rare occasions when L'Hospital's Rule is your friend.

We cannot prove this without some assumptions on $f,u,v$: I am going to assume that all the derivatives I mention exist and are continuous.

Let $$H(h):=f(u(t+h),v(t+h))-f(u(t),v(t+h)).$$

To calculate $\lim_{h\to 0}\frac{H(h)}{h}$ it suffices to calculate $\lim_{h\to 0}H'(h)$, provided this exists. That is we want to calculate the limit as $h\to 0$ of $$ f_{u}(u(t+h),v(t+h))u'(t+h)+f_{v}(u(t+h),v(t+h))v'(t+h)\\ -f_{u}(u(t),v(t+h))\cdot 0-f_{v}(u(t),v(t+h))v'(t+h). $$

Now we can use the (assumed) continuity and take the limit to get $$ f_{u}(u(t),v(t))u'(t)+f_{v}(u(t),v(t))v'(t)\\ -f_{u}(u(t),v(t))\cdot 0-f_{v}(u(t),v(t))v'(t) $$ which is just $$ f_{u}(u(t),v(t))u'(t). $$

The other limit also evaluates to $f_{u}(u(t),v(t))u'(t)$: either use the definition of partial derivative (if it indeed applies) or argue in the same way as I have done.

ancient mathematician
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