Consider a function $A: R^n \to R^m$ (smooth) and another $v: R \to R^n$ defined as $v: R \ni u \mapsto u \cdot w $ for $w\in R^n$. Finally let $x \in R^m$.

Now define a new function $B : R \to R$ $$ u \mapsto A (v(u)) ^t x. $$

We would like to say something clever about the derivatives $$ \frac{\partial^k B}{\partial u ^k} . $$ If $n=m=1$ then by the chain rule $$ \frac{\partial^k B}{\partial u ^k} = \frac{\partial^k A}{\partial u ^k} (u w ) w^k x $$

And I'm wondering if something similar exists in general i.e. if $\tfrac{\partial^k B}{\partial u ^k}$ can be expressed as some sort of linear combination of mixed partial derivatives of $A?$ (and powers of $w$). I'm comfortable with multiindices.

What if $m=n$ or $m=1$?


Use the multi-variable chain rule: if $f \colon \Bbb{R}^k \to \Bbb{R}^m$ and $g \colon \Bbb{R}^m \to \Bbb{R}^n$, then for $v \in \Bbb{R}^k$, then $d(g \circ f)_v$, the Jacobian matrix of $g \circ f$ at the point $v$, is given by:

$$ d(g \circ f)_v = dg_{f(v)} \circ df_v $$

In your case we can think of this as a three-step composition. In your notation from above, let's call $C \colon \Bbb{R}^m \to \Bbb{R}$ the map which takes the inner product with the fixed vector $x$. So for $y \in \Bbb{R}^m$, $C(y) = \left<y, x\right>$. Or maybe $C(y) = y^tx$ in the notation you use above. Then your function $B$ is the three-step composition:

$$ B = C \circ A \circ v $$

I'm not sure what you mean by the "partial derivatives" of $B$, though. It is just a function from $\Bbb{R}$ to $\Bbb{R}$, so there is just the one ordinary single-variable derivative $\frac{dB}{du}$. We can express this derivative as a composition using the chain rule:

$$ \frac{dB}{du} = dC_{A(v(u))} \circ dA_{v(u)} \circ dv_u $$

Unfolding notation, you can see that $dv_u$ is just the vector $w$ (written as a column vector), $dA_{v(u)}$ is the jacobian matrix of $A$ (with entries $\frac{\partial A_i}{\partial x_j}$) at the point $uw$, and $dC_{A(v(u))}$ is the vector $x$ (written as a row vector). So taking all this into account, we can re-write the chain rule above in this case as

$$ \frac{dB}{du} = \left< x, dA_{uw} w\right> = x^t dA_{uw} w $$

If you want, you can write this out with indices by doing the matrix multiplication. Let's call $y_1,\dots,y_n$ the coordinates on $\Bbb{R}^n$. Then this would be:

$$ \frac{dB}{du} = \sum_{i=1}^m \sum_{j=1}^n x_i w_j \frac{\partial A_i}{\partial y_j}(uw) $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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