# Partial Derivatives and Linear Map

Proposition:

Suppose that $T : R^n \to R^m$ is the linear transformation defined by

$T(x) = Mx$

for some m × n matrix $M$. Then

$DT(x) = M$

for all points $x \in R^n$

where $D$ is the partial derivative matrix. (Jacobian?)

Question:

I don't understand what is being said. $T(x)$ is a linear transformation on $x$. How does the partial derivative of $T(x)$ lead to the transformation matrix. Neither do I have an algebraic intuition nor a geometric one.

Further, How is the total derivative of $g(x,y,z)$ equal to $Dg(x,y,z) \begin{pmatrix} x \\ y\\ z \end{pmatrix}$?

This is stated without proof. There is a chance, I made a wrong interpretation so I am pasting the portion of the text where it appears.

Is it that the change in $x$ in all dimensions of the output of $g(x,y,z)$ multiplied by $x$ and similar for y and z gives a total derivative. I don't seem to understand.I know the total derivative is a derivative taking into account that other variables are not constant during differentiation by one variable.

-
Regarding the second part: Confusingly, there are many things which are given the name "total derivative". But this doesn't appear to match up with any of them. Are you sure of the transcription? It could be that $x, y, z$ are functions of a single variable $t$, and the matrix on the right could be $\begin{pmatrix}x'(t) \\ y'(t) \\ z'(t)\end{pmatrix}$. Then this would just be the chain rule. – Dylan Moreland Feb 21 '12 at 15:08
@DylanMoreland, I have added a picture of the part where it appears. – Inquest Feb 21 '12 at 15:14
Thanks for scanning; that really helps us. [It would be good to say what $g$ and $f$ are, too.] I would not use the same letters for $(1, 0)$ (the point at which you're finding the total derivative; maybe write a general point as $(a, b)$) and the vector $(x, y)$ on which you are evaluating the derivative. – Dylan Moreland Feb 21 '12 at 15:16
@DylanMoreland, I added the whole problem. – Inquest Feb 21 '12 at 15:19

For algebraic intuition: The derivative of a function is the best linear approximation to it. In the case when the function is linear already, it's its own best linear approximation, hence it's its own derivative.

As a reminder, for multivariate $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$, the derivative $Df(x)$ at the point $x\in \mathbb{R}^m$ is defined to be a linear map $A:\mathbb{R}^m\rightarrow\mathbb{R}^n$ satisfying

$$\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x) - Ah}{\|h\|}=0$$

It's not too hard to show that if $A$ exists, it's unique.

Now, we simply check that for $T(x) = Mx$, everything works.

$$\lim_{h\rightarrow 0} \dfrac{T(x+h)-T(x) - Mh}{\|h\|} = \lim_{h\rightarrow 0} \dfrac{M(x+h)-Mx-Mh}{\|h\|} = \lim_{h\rightarrow 0}\dfrac{Mx+Mh-Mx-Mh}{\|h\|}=0$$ since the numerator is $0$.

In answer to the comment, the derivative at a point is the best linear approximation at that point (though we think of the point as being shifted to the origin). Thus, in your example of $f(x) = x^3$, the derivative, $3x^2$, at a point (say $x=2$) gives a slope of $12$. The linear approximation this corresponds to is $12x$.
In fact, this issue is one it took me a long time to come to terms with. For functions $f:\mathbb{R}\rightarrow\mathbb{R}$, we think of the derivative of the whole function as a new function $f'(x)$. With more variables, we change view point: The derivative of a function at a point is itself a function (linear transformation). The relationship between the 2 is that a linear transformation from $\mathbb{R}\rightarrow\mathbb{R}$ can be identified with a single real number $\mathbb{R}$ via a pretty canonical choice of basis. The derivative of a function $f:\mathbb{R}\rightarrow\mathbb{R}$ at a point $p$ is the linear map "multiply by $f'(p)$".
1. In the definition of the derivative, should it be $f(x+h) - f(x)$ instead of $f(x+h) - f(h)$ or am I missing something. 2. I understood the interpretation but I can't seem to completely appreciate the fact that a derivative is a linear approximation. How is $3x^2$ a linear approximation to $x^3$? – Inquest Feb 21 '12 at 15:39