Proof of multivariable chain rule for $f(t,x(t))$

Question

I'm trying to prove the multivariable chain rule for this base case:

$$f(t,x(t))$$

Here's what I tried:

$$\lim_{h\to 0} \frac{f(t+h, x(t+h)) - f(t,x(t))}{h} = \lim_{h\to 0} \frac{f(t+h, x(t+h)) - f(t,x(t))}{x(t+h)-x(t)}\frac{x(t+h)-x(t)}{h} = $$ $$ \lim_{h\to 0} \frac{f(t+h, x(t+h)) - f(t, x(t+h))+f(t, x(t+h))- f(t,x(t))}{x(t+h)-x(t)}\frac{x(t+h)-x(t)}{h} = $$

$$\lim_{h\to 0}\frac{f(t+h, x(t+h)) - f(t,x(t+h))}{x(t+h)-x(t)}\frac{x(t+h)-x(t)}{h} +$$ $$\lim_{h\to 0}\frac{f(t,x(t+h))-f(t,x(t))}{x(t+h)-x(t)}\frac{x(t+h)-x(t)}{h} = $$

I can't view these things as partial derivatives of $f$. Can somebody help me?

Answer 1

You can use $f(t+h,x(t+h))-f(t,x(t+h))=hf_x(t+\theta_1 h,x(t+h))$ for some $\theta_1\in(0,1)$ and $f(t,x(t+h))-f(t,x(t))=f(t,x(t)+hx'(t+\theta_2h))-f(t,x(t))=hx'(\theta_2h)f_y(t,x(t)+\theta_3h\theta_2x'(t+\theta_2h)$.

Answer 2

I'm gonna give you a different perspective. It might help you, as long as you're familiar with "little-o" notation.

First, let's rewrite a derivative as a local linear (affine to be pedantic) approximation. More on this here. $$ x(t+h) = x(t)+h\,x'(t)+o(h) \tag1$$

As you can easily see, if you isolate $x'(t)$, you go back to the usual definition: $$ x'(t)=\frac{x(t+h)-x(t)}{h}+o(1) $$

Notice that we're assuming $h$ arbitrarily small, as you do when you take the limit for $h\to0.$

You can do the same for partial derivatives: \begin{align} f(t+h,x)&=f(t,x)+h\,f_t(t,x)+o(h) \tag2\\ f(t,x+k)&=f(t,x)+k\,f_x(t,x)+o(k) \tag3 \end{align}

Now let's see what we can do for the total derivative: \begin{align} f\left(t+h,\,x(t+h)\right) &= f\left(t+h,\,x(t)+h\,x'(t)+o(h)\right) & &\text{(1)}\\ &=f(t+h,x)+k\,f_x(t+h,x)+o(k) & &\text{(3) with }k=h\,x'(t)+o(h)\\ &=f(t+h,x)+\left[h\,x'(t)+o(h)\right]f_x(t+h,x)+o(h) & &\text{in this case }o(k)=o(h)\\ &=f(t,x)+h\,f_t(t,x)+\left[h\,x'(t)+o(h)\right]f_x(t+h,x)+o(h) & &\text{(2)} \end{align}

Notice that for ease of notation I wrote $x$ meaning $x(t)$. Since we're dealing only with first order derivative we can safely ignore higher order ones: $$ f_x(t+h,x)=f_x(t,x)+h\frac{\partial}{\partial t}f_x(t,x)+o(h)=f_x(t,x)+o(1)$$

Thus \begin{align} f\left(t+h,\,x(t+h)\right) &=f(t,x)+h\,f_t(t,x)+h\,x'(t)\,f_x(t,x)+o(h)\\ &=f(t,x)+h\underbrace{\left[f_t(t,x)+x'(t)\,f_x(t,x)\right]}_{\frac{d}{dt}\,f\left(t,x(t)\right)}+o(h) \end{align}

Hence $$ \frac{d}{dt}\,f\left(t,x(t)\right) = f_t(t,x)+x'(t)\,f_x(t,x)$$

I know this isn't exactly what you asked for, but I prefer this notation rather than using the standard definition. I think this is more clear on what becomes a partial derivative and what vanishes because is infinitesimal of higher order.

Answer 3

You don't have to "prove the chain rule", but to find out what it has to say in the particular case described in the question.

We are given a function $(u_1,u_2)\mapsto f(u_1,u_2)$ of two variables $u_1$, $u_2$, and a function $t\mapsto x(t)$. These givens are then combined to the function $$\phi(t):=f\bigl(t,x(t)\bigr)\ ,$$ and you are required to compute $\phi'(t)$. The chain rule says that $$\phi'(t)=f_{.1}\bigl(t,x(t)\bigr)u_1'(t)+f_{.2}\bigl(t,x(t)\bigr)u_2'(t)=f_{.1}\bigl(t,x(t)\bigr)\cdot1+f_{.2}\bigl(t,x(t)\bigr)x'(t)\ .$$