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From An Introduction to Manifolds by Tu:

Let $g(t) = f(p + t (x-p))$ where $x,p \in \Bbb R^n$ and $t \in \Bbb R$.

Then $\frac{d}{dt}f(p + t(x-p)) = \sum(x_i-p_i)\frac{\partial f}{\partial x_i}(p + t(x-p))$

How is this derivative derived? Where does the sum come from and all the other partial derivatives?

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Oliver G
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2 Answers2

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This is about the multivariable chain rule. Here I'll take $n=2$ as an example to explain the derivative equation. For convenience, I assume the derivatives of $f$ is continuous, although this assumption is not neccessary. Actually, the differentiability of $f$ will be enough, see here for example.

In this case, $g(t)=f(p_1+t(x_1-p_1),p_2+t(x_2-p_2))$. Thus \begin{align*} \frac{g(t)-g(t_0)}{t-t_0}&=\frac{f(p_1+t(x_1-p_1),p_2+t(x_2-p_2))-f(p_1+t_0(x_1-p_1),p_2+t_0(x_2-p_2))}{t-t_0}\\&=\frac{f(p_1+t(x_1-p_1),p_2+t(x_2-p_2))-f(p_1+t_0(x_1-p_1),p_2+t(x_2-p_2))}{[p_1+t(x_1-p_1)]-[p_1+t_0(x_1-p_1)]}\cdot(x_1-p_1)\\&+\frac{f(p_1+t_0(x_1-p_1),p_2+t(x_2-p_2))-f(p_1+t_0(x_1-p_1),p_2+t_0(x_2-p_2))}{[p_2+t(x_2-p_2)]-[p_2+t_0(x_2-p_2)]}\cdot(x_2-p_2). \end{align*} Letting $t\to t_0$ and using the continuity of $f'_2$ gives the desired result.

Feng
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  • So when the author writes: $\frac{d}{dt}f(p + t(x-p)) = \sum(x_i-p_i)\frac{\partial f}{\partial x_i}(p + t(x-p))$ is he abusing notation? Since if $f = f(g_1(t), \dots, g_n(t))$ then $\frac{d}{dt}f = \sum(x_i-p_i)\frac{\partial f}{\partial g_i}( p+ t(x-p))$. Equating the author's notation with the definition we have $g_i(t) = x_i = p_i + t(x_i - p_i)$, but this doesn't make any sense. – Oliver G Aug 03 '19 at 17:03
  • @OliverG You are right. Here $\frac{\partial f}{\partial x_i}$ means the partial derivative with respect to the i-th variable, and I prefer to write it as $f_i'$. – Feng Aug 03 '19 at 22:35
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The general formula is $\frac d {dt} f(g(t,x))=\sum_{i=1}^{n} \frac {\partial } {\partial x_i} f(g(t,x) \frac {\partial} {dt} g(t,x)$. This is just the chain rule.

Kavi Rama Murthy
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