Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let us consider $g: \mathbb R^n \to \mathbb R^m$ , $f:\mathbb R^m \to \mathbb R^k$ be $C^1$ functions . Define $F= f\circ g$. I am having the problem in finding the partial derivative of $F$,

$$\frac {\partial F}{\partial x_i} = \frac{\partial \{f_1(g(x)), .......f_k(g(x)) }{\partial x_i}$$

Let $g(x)=y \in \mathbb R^m$ which is not constant. $$\frac {\partial F}{\partial x_i} = \frac{\partial \{f_1(y), .......f_k(y)}{\partial y} . \frac{\partial y}{\partial x_i}$$

$$=\nabla f_{(g(x))} \cdot \sum_{i=1}^{m} \frac{\partial g_i(x)}{\partial x_i} e_i$$ This doesn't make sense , because the first one has $k$ components and the second term has $m$ components . Can someone point out where i am going wrong ?

share|cite|improve this question
up vote 2 down vote accepted

$F:\mathbb R^n\rightarrow \mathbb R^k$ has $k$-components. Then

$$\frac{\partial F_r}{\partial x_i}(x)=(\text{chain rule})= \sum_{j=1}^m\frac{\partial f_r}{\partial y_j}(g(x))\frac{\partial y_j}{\partial x_i}, $$

for all $r\in \{1,\dots,k\}$, where $x=(x_1,\dots,x_n)$ and $y=g(x)\in\mathbb R^m$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.