Can I take the gradient of a function like this?

Question

Let us say I have a vector function

$$r(u,v) = \langle u,v,1000+u^2+v^2 \rangle.$$

I also have,

$$p(x,y,z) = 5e^{(x + y + z^2)}.$$

Let, $Q(u,v) = (P \circ r)(u,v).$ Then,

$$Q(u,v) = 5e^{(u+v+(1000+u^2+v^2)^2)} $$

Would $$\frac{\partial Q}{\partial u} = \frac{\partial}{\partial u} (5e^{(u+v+(1000+u^2+v^2)^2)}) ? $$

Answer 1

It is correct. You could also use the chain rule. Denoting $r(u,v) = \bigl(r_1(u,v), r_2(u,v),r_3(u,v)\bigr)$: $$ \frac{\partial Q}{\partial u} = \frac{\partial r_1}{\partial u} \frac{\partial P}{\partial z}\bigl(r(u,v)\bigr) + \frac{\partial r_2}{\partial u}\frac{\partial P}{\partial y}\bigl(r(u,v)\bigr)+ \frac{\partial r_2}{\partial u}\frac{\partial P}{\partial z}\bigl(r(u,v)\bigr) = P\bigl(r(u,v)\bigr)\bigl(1+4(1000+u^2+v^2)\bigr). $$

Answer 2

According to the chain rule, the Jacobian matrix (matrix of derivatives) of $F=f\,o\,g\,:\mathbb{R}^n\rightarrow\mathbb{R}^m$, where $f:\mathbb{R}^p\rightarrow\mathbb{R}^m$ and $g:\mathbb{R}^n\rightarrow\mathbb{R}^p$ is defined as:

$$(Df)(\textbf{x})=(D(f\,o\,g))(\textbf{x})=(Df)(g(\textbf{x}))(Dg)(\textbf{x})$$

Therefore, in this case:

$$(DQ)(u,v)=(D(P\,o\,r))(u,v)=(DP)(r(u,v))(Dr)(u,v)$$

We have:

$$DP(x,y,z)=\begin{pmatrix}\frac{\partial P}{\partial x} & \frac{\partial P}{\partial y} & \frac{\partial P}{\partial z}\end{pmatrix}\Rightarrow (DP)(r(u,v))=DP(r_1,r_2,r_3)=\begin{pmatrix}\frac{\partial P}{\partial r_1} & \frac{\partial P}{\partial r_2} & \frac{\partial P}{\partial r_3}\end{pmatrix}$$

$$Dr(u,v)=\begin{pmatrix}\frac{\partial r_1}{\partial u} & \frac{\partial r_1}{\partial v} \\ \frac{\partial r_2}{\partial u} & \frac{\partial r_2}{\partial v} \\ \frac{\partial r_3}{\partial u} & \frac{\partial r_3}{\partial v} \end{pmatrix}$$

Where $r(u,v)=(r_1(u,v),r_2(u,v), r_3(u,v))$

But also by definition:

$$DQ(u,v)=\begin{pmatrix}\frac{\partial Q}{\partial u} & \frac{\partial Q}{\partial v} \end{pmatrix}$$

Applying the chain rule:

$$DQ(u,v)=\begin{pmatrix}\frac{\partial P}{\partial r_1} & \frac{\partial P}{\partial r_2} & \frac{\partial P}{\partial r_3}\end{pmatrix} \cdot \begin{pmatrix}\frac{\partial r_1}{\partial u} & \frac{\partial r_1}{\partial v} \\ \frac{\partial r_2}{\partial u} & \frac{\partial r_2}{\partial v} \\ \frac{\partial r_3}{\partial u} & \frac{\partial r_3}{\partial v} \end{pmatrix}$$

Then, we know:

$$\begin{pmatrix}\frac{\partial Q}{\partial u} & \frac{\partial Q}{\partial v} \end{pmatrix}=\begin{pmatrix}\frac{\partial P}{\partial r_1} & \frac{\partial P}{\partial r_2} & \frac{\partial P}{\partial r_3}\end{pmatrix} \cdot \begin{pmatrix}\frac{\partial r_1}{\partial u} & \frac{\partial r_1}{\partial v} \\ \frac{\partial r_2}{\partial u} & \frac{\partial r_2}{\partial v} \\ \frac{\partial r_3}{\partial u} & \frac{\partial r_3}{\partial v} \end{pmatrix}$$

And by multiplying for the first column on the right of the expression:

$$\frac{\partial Q}{\partial u}=\frac{\partial P}{\partial r_1} \cdot \frac{\partial r_1}{\partial u}+ \frac{\partial P}{\partial r_2} \cdot \frac{\partial r_2}{\partial u}+ \frac{\partial P}{\partial r_3} \cdot \frac{\partial r_3}{\partial u}$$

And for the second column:

$$\frac{\partial Q}{\partial v}=\frac{\partial P}{\partial r_1} \cdot \frac{\partial r_1}{\partial v}+ \frac{\partial P}{\partial r_2} \cdot \frac{\partial r_2}{\partial v}+ \frac{\partial P}{\partial r_3} \cdot \frac{\partial r_3}{\partial v}$$

Then, the gradient of the function is:

$$\vec{\nabla} Q(u,v) = (DQ(u,v))^T=\begin{pmatrix}\frac{\partial P}{\partial r_1} \cdot \frac{\partial r_1}{\partial u}+ \frac{\partial P}{\partial r_2} \cdot \frac{\partial r_2}{\partial u}+ \frac{\partial P}{\partial r_3} \cdot \frac{\partial r_3}{\partial u} \\ \frac{\partial P}{\partial r_1} \cdot \frac{\partial r_1}{\partial v}+ \frac{\partial P}{\partial r_2} \cdot \frac{\partial r_2}{\partial v}+ \frac{\partial P}{\partial r_3} \cdot \frac{\partial r_3}{\partial v} \end{pmatrix}$$

This takes simpler computations for very complex functions or compositions of a large number of functions. Of course, in the case posed by the OP the composition may not be so complex as to require this method. You can simply find the explicit representation of the function $Q$ and then find the partial derivatives from there.