How To Understand Derivatives: The Quotient Rule, Exponents, and Logarithms

Last time we tackled derivatives with a "machine" metaphor. Functions are a machine with an input (x) and output (y) lever. The derivative, dy/dx, is how much "output wiggle" we get when we wiggle the input:

This is a companion discussion topic for the original entry at

Thanks so much for the explanation. Really help a lot! 1 Q still confused regarding quotient rule (dg part):

If we used directly g+dg instead of x+1 in the calculation:
1/(g+dg) - 1/g
= (g-g+dg) / (g^2+gdg)
= -dg / (g^2 + g

May I ask why is the g*dg ignored or cancelled? Is there an intuitive reasoning for it? Thanks so much in advance.

hoping to get more example to solve

@Joe: I hear you – we slice and dice concepts and miss the cohesive whole. All the calculus rules are just examples of how different subparts can contribute to the whole, but I’m only seeing that now, 10+ years after high school. Ugh.

And yeah – there’s so much “don’t do this, I don’t know why, but don’t!” in math. Why is it against the rules? What are the “rules”? Limits are a seatbelt introduced to address theoretical concerns many, many years after Calculus was put into use. Learning about seatbelts is fine, but don’t dive into them before you explain what a car [i.e., calculus] is!

you are a geneous

When will Math curriculums begin combining concepts in meaningful ways like this? Calculus classes like to split ‘Power Rule,’ ‘Quotient Rule,’ and ‘Chain Rule’ into discrete sections, when really they’re consequences of the same basic idea. Perhaps it’s less labor-intensive teaching distinct formulas to be memorized, but it’s just another reason people hear ‘Calculus’ and immediately glaze over.

And while I’m lamenting–your mention of infinitesimals brings up another sore spot of mine. A Calc TA told me how separating ‘dy/dx’ is ‘against the rules,’ as you say, and I took it to heart. Imagine poor, confused me a couple semesters later in DiffEq: “I thought this was against the rules!” The limit-based approach to teaching Calculus needs some serious revision, particularly for non-mathematicians moving into practical fields.

Very informative and great analogy :slight_smile:

@Hitoshi: Thanks for the comment! Just fixed up the article :).

As usual, a nice article.

By the way, in section Division (Quotient Rule), there is an extra ‘]’ at the end of the line
. f changes by df, contributing area df * m = df * (1 / g)]

Thanks for the great article.

@Kinar: Thank you!

What helped me understand derivatives is this: If y=e^x, then y’=e^x, which is of course related to your favorite number, e, which does seem to have more significance than pi. A graph and an explanation could help others.

@Gulrez: Hah, thanks for the kind words :). I’m hoping to do more on a bunch of math topics, appreciate the support!

u are genius!what will be ur next topic ? I eagerly wait for ur math posts.

Thanks Tim, a follow-up charting the path of e^x would be a good idea.

If I had known this existed while I was taking calculus, there would have been so fewer headaches. I had always known on a subconscious level that there were connections between calculus and the earlier maths–my teacher even confirmed that by joking that all other classes were “pre-calculus”–but for the life of me, I could never find those connections. And they were right there mocking me the whole time! This helped more than any lecture, peer teaching, or textbook ever could. Thanks!

i like how u sort out but the side of Q u nid to expand a little bit so that we gain better and best…thump up khalid and i ll kip following ol

Thanks Jackson, great question. Intuitively, think about taking a “single step forward”, which is 1*dx. Another way of seeing it: when taking the derivative, we split our continuous function into discrete steps (a single dx wide at each step) and see our rate of change when we increment by the next dx.

An analogy: we represent a photo with individual pixels (dx) and step through one pixel at a time. The pixels are chosen at an “infinitely small retina resolution” where we don’t notice them at the macro scale. (There’s more on limits later in this series.)

Thank you for the time you’ve put into these articles they’ve helped me a lot and I’m glad to know there are people who care about intuition and share it, but I’m confused about your intuition of the natural log. Why is the derivative always predicting the next increment by one? Why not .5? Shouldn’t it be infitestimally small because it is using the input of the a naturally growing function?

Sorry for my inconvienience but I’m confused how you got 1/x. Wouldn’t the derivative be dx/x because dx would be the change and x would be the current value as dx approach 0.I’m just confused why 1=dx instead of approaching 0.

No worries, great question, I realize it can be unclear. I start with scenarios where “dx = 1” (which is a GIANT step) to estimate results in my head. Then, I can set dx = 0 (taking the limit) to get an exact prediction.

Let’s say I want the derivative of x^2. I imagine going from 10^2 to 11^2 (we jumped from x=10 to x=11, so dx=1). The difference is 21, or 2x + dx (20 + 1). I can then set dx = 0 and get the exact answer of 2x. (If there was no gap between x and the next value, the derivative would be 2x.)

The natural log is harder to compute: it’s the time e^x needs to grow from 1 to x. How does it change?

Imagine going from 10 to 11 (again, dx=1). Here, we’re at 10 and we grow exponentially up to 11. Since e^x assumes we’re growing at 100% of our current value, it takes 1/10 of a unit time to get to 11. (10 + (1/10)*10 = 11).

Now, this isn’t quite accurate because as we’re going to 11, we’re getting faster. I.e., when we’re at 10.5 we’re growing at 10.5 units per unit time, not the 10 we expected. Removing the imaginary dx fixes this (we assume there is no midpoint between x=10 and the next value, so it really is a perfect 1/x amount of time we wait).