How do derivatives even work?

Question

I'm a student currently looking into partial derivatives and was writing notes for a simple example. $$\require{cancel}\text{find the derivative with respect to x:}\\x^2+xy-y^2=8$$ I got to the point where I was differentiating the $-y^2$ term, I wrote $\frac{d(-y^2)}{dy}\frac{dy}{dx}$. This got me thinking, with my low level knowledge of what derivatives even are. Why does it not work out when I take out a factor of y and cancel terms like this $$-y\frac{\cancel{dy}}{\cancel{dy}}\frac{dy}{dx}$$ Any knowledge and deep understanding is really appreciated!

Answer 1

(This is more a comment than an answer, as it's asking for clarification, but it's too long for a comment.)

Is that the full text and context of an example that was assigned to you? Because I would describe that as "missing a boat-load of context".

See, there are (at least) two ways to view that formula.

1. Define $f(x,y) = x^2+ xy - y^2$. Differentiate $f$ w.r.t. $x$.

This version would use the notation for partial derivatives (which you mentioned), but that's $\frac{\partial f}{\partial x}$, not $\frac{dy}{dx}$, which is what you've written. Although just what the "$=8$" would be doing in there in this case I'm not sure. And in this case, $\frac{\partial (y^2)}{\partial x}$ is actually equal to $0$, as $y$ doesn't depend on $x$!

2. Assume the equation $x^2+ xy - y^2 = 8$ defines $y$ implicitly in terms of $x$. Compute $\frac{dy}{dx}$.

In this version you would think of $y$ as a function of $x$, i.e. $y(x)$, and it should be clear that in this case you can't "factor $y$ out" anymore than your could, if, say, $g(x)$ were a function of $x$, write $$\frac{d(g\cdot g)}{dx} = g\cdot \frac{d g}{dx}$$

Distinguishing between versions 1 and 2 depends on making clear whether we are in a situation where $x$ and $y$ can vary freely and independently, or one where a change in $x$ induces a change in $y$. The instruction "find the derivative with respect to $x$" does nothing to help us understand which of the situations we are in, and the use of the description "partial derivatives" and the notation $\frac{dy}{dx}$ point in completely opposite directions. This example needs more details and more context to be able to provide an answer.

Answer 2

There's several important facts about derivatives that might help untangle the mess:

1. df/dx is linear operator.

2. f is not a function, but polynomial

3. to transform derivatives to your ordinary linear algebra, you just need to represent your polynomial in basis [1,x,x²,x³,x⁴,... ], and the coefficients need to be separable.

4. polynomial representation in matrix form is just: $$\begin{bmatrix}a& b& c& d& e\\f& g& h& i& j \\k &l &m& n& o\\p& q& r& s& t\\u& v &w &x &y\\ \end{bmatrix} \begin{bmatrix}1\\x\\x²\\x³\\x⁴\end{bmatrix}$$

5. Derivative in matrix form is just(for size 5): \begin{bmatrix}0&0&0&0&0\\1&0&0&0&0\\0&2&0&0&0\\0&0&3&0&0\\0&0&0&4&0\end{bmatrix}

I.e. it's an identity matrix which has been "shifted" a little with coefficients correctly setup.

This uses the normal polynomial derivative calculation: $$f(x)=x^n \Rightarrow f'(x)=n x^{n-1}$$ and ability to move constants outside of derivative $$f(x)=c g(x) \Rightarrow f'(x) = c g'(x)$$ and of course the sum rule $$f(x)=g(x)+h(x) \Rightarrow f'(x)=g'(x)+h'(x)$$

6. To handle multiple variables like x and y, you might need to change the basis from [1,x,x²,x³,x⁴] to [1,y,y²,y³,y⁴]. Linear algebra has tons of information how the basis change can be done. Choosing correct basis for your derivative makes the derivative matrix look reasonable/easier to remember.
7. derivative matrix anyway only works if your calculations are in correct basis.