Does Thompson's Calculus Made Easy have any discussion of implicit differentiation?

Question

I didn't see it in the index but would like to read his treatment if there is one. The version I'm thinking of is here: http://calculusmadeeasy.org/

(Sorry if this isn't an appropriate question for this forum, but I thought it might be alright.)

ADDITIONAL EDIT: Page 2 of this article has a differential-based treatment of implicit differentiation, although I don't quite understand it, nor am I sure that it's identical to how Thompson treated it: http://math.oregonstate.edu/bridge/papers/differentials.pdf.

Answer 1

I think B. Nunez gave a great answer, but I will follow on with a more specific example. It's been a long time since I read Calculus Made Easy (CME), so apologies if this doesn't fit their methodology exactly.

In "traditional" calculus, you find the derivative with respect to a particular variable. In CME, you find the derivative by incrementing each variable according to its own increment. So, you replace $x$ with $x + dx$ and $y$ with $y + dy$.

So, let's say you have:

$$x^2 + y^2 = 1.$$

The CME approach would have you replace each $x$ with $x + dx$ and $y$ with $y + dy$. This then becomes:

$$ (x + dx)^2 + (y + dy)^2 = 1 \\ x^2 + 2x\,dx + dx^2 + y^2 + 2y\,dy + dy^2 = 1 \\ 2x\,dx + dx^2 + 2y\,dy + dy^2 = 0 \\ 2x\,dx + 2y\,dy = 0 \\ 2y\,dy = -2x\,dx \\ \frac{dy}{dx} = -\frac{x}{y} $$

The simplification from step 2 to step 3 is because we subtracted the original equation from both sides. The simplification from step 3 to step 4 is because the square of an infinitesimal is infinitely smaller than the original infinitesimal, and therefore infinitely more insignificant. Then I just solved for $\frac{dy}{dx}$.

If you are interested, the way that I teach implicit differentiation is here.

Answer 2

Thompson's approach to calculus is through the use of differentials, and thus his treatment of the derivative is essentially the same as what one would call "implicit differentiation" in standard textbooks: given an equation that relates $y(x)$ and $x$, differentiate all terms with respect to $x$ to obtain an implicit definition of $dy/dx$. Consequently, there's no need to distinguish between finding the derivative of a function and takes derivatives on an implicit expression. See e.g. http://calculusmadeeasy.org/6.html