0
$\begingroup$

Say we have a function $$f:X\times Y \to \mathbb{R}$$ where $Y\subseteq \mathbb{R}^n$. I want to calculate $\frac{\partial}{\partial t}f(x,y_0+tu_i)$ for a fixed $y_0 \in Y$ and $u_i\in \mathbb{R}^n$ is the i'th standard basis vector for a fixed $i$ and $t\in \mathbb{R}$. Now, just using the chain rule I think we get $$\frac{\partial}{\partial t}f(x,y_0+tu_i)=\frac{\partial f}{\partial y}(x,y_0+tu_i)u_i$$ But we haven't assumed that $f$ is differentiable. We only assume the existence of the partial derivative of $f$ w.r.t the $i$'th coordinate of $y$, so is the above even well-defined?

$\endgroup$

1 Answer 1

1
$\begingroup$

The important point is not to use the chain rule (since you don't know $f$ is differentiable), but just the definitions. If you're calculating this partial derivative with respect to $t$, you have fixed $x=x_0$ as well. Note that if we set $t=s+t_0$, we have $$\frac d{dt}\Big|_{t=t_0} f(x_0,y_0+tu_i) = \frac d{ds}\Big|_{s=0} f(x_0,(y_0+t_0u_i)+su_i) = \frac{\partial f}{\partial y_i}(x_0,y_0+t_0u_i).$$ (The final equality is just the definition of the partial derivative.)

If you don't know that $f(x_0,\cdot)$ is a differentiable function (of $y$), then what you've written doesn't actually make sense.

$\endgroup$
2
  • $\begingroup$ I think I understand, but just to be sure: By definition, the partial derivative is $$\lim_{h\to 0} \frac{f(x_0,y_0+(t_0+h)u_i)-f(x_0,y_0+t_0u_i)}{h}$$ $$=\lim_{h\to 0} \frac{f(x_0,y_0+t_0u_i+hu_i)-f(x_0,y_0+t_0u_i)}{h}$$ $$=\frac{\partial f}{\partial y_i}(x_0,y_0+t_0u_i)$$ Because the second expression is the definition of the partial derivative w.r.t $y_i$? $\endgroup$
    – Snildt
    Commented Jan 22, 2022 at 19:44
  • $\begingroup$ Yes, that's correct. $\endgroup$ Commented Jan 22, 2022 at 19:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .