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In the book Structure and Interpretation of Computer Programs, there is an interesting example on how one might explore symbolic data in programming. They used the differentiation algorithm. That is, there is an algebraic expression (E.G. x+3) and find the derivative of this expression with respect to x.

Now I've read and understood the program they created, but unfortunately have no idea what I'm actually doing. So, I began exploring derivatives because I thought I might learn something amazing.

Unfortunately, all the explanations I found were in terms of slopes, y-intercept, x-intercept and all that. My eyes are destroyed, so I can't imagine those slope things and graphing function stuff, and no one explained to me what those things are in a clear way before.

Is there an explanation of derivatives and how you might solve for one without using visual slopes?

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Derivatives are rates of change at an instant.

If a car goes $60$ miles in $2$ hours, then its average speed during that time is $\dfrac{60} 2 = 30$ miles per hour. But how fast is it going at a particular instant? It goes $0$ miles in $0$ hours during that instant, regardless of what its speed is. Overcoming that difficulty is what derivatives do. For example, suppose $y=x^3$. When $x=2$ then $y=8$ and $\left.\dfrac{dy}{dx}\right|_{x=2} = 12$. That means that when $x=2$ and $y=8$ then $y$ is changing $12$ times as fast as $x$ is changing.

Here's one way to make sense of the car's speed at an instant. Suppose the speed of car A may vary, but car B is moving at a constant speed of $30$ miles per hour in the same direction through all eternity. If car A overtake car B at a particular instant, that does not mean car A is going faster than car B at that instant, because what could be happening is this: car A is going faster than car B as car A approaches car B from behind, but car A is gradually getting slower, until the instant they meet, when car A has slowed down to exactly $30$ miles per hour, but then car A starts gaining speed again and overtakes car B. But car A's overtaking car B does mean car A's speed at that instant is not slower than $30$ miles per hour. Thus we can say car A is not moving more slowly than some cars whose speed is constant, and is not moving faster than other cars whose speed is constant. There is a boundary between those speeds that are faster than car A's speed at that instant, and those that are slower. That boundary is car A's speed at that instant.

That is the value of a derivative at that instant.

  • this is beautiful, thanks. In your example, when x is 2 and y becomes 8, how does y change 12 times faster than x? And in the cars example, can there be many derivatives as many as car a's speed changes? – lightning_missile Jan 18 '16 at 17:47
  • @morbidCode : If $y=x^3$ then $dy/dx = 3x^2$ is when $x=2$ that is $3\cdot2^2=12$. I'm not sure I understand your second question. Car A's speed is the derivative of the distance it's traveled with respect to time. The value of the derivative would typically be different at different times. ${}\qquad{}$ – Michael Hardy Jan 18 '16 at 17:56
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The geometry all comes down to this equation:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}.$$

The quantity that we're taking the limit of can be understood as:

  1. The slope of the line between the points $(x,f(x))$ and $(x+h,f(x+h))$
  2. The rate of change of $f$ on the interval $[x,x+h]$.

After we take the limit, we have

  1. The slope of the tangent line to the graph of $f$ at the point $(x,f(x))$.
  2. The rate of change of $f$ at the point $x$.

That last thing is a bit strange if you think about it carefully: what does the rate of change of something at a single point mean? That is part of why we talk about these other interpretations.

Another interpretation is that the tangent line at $x$ is the best approximation to $f$ by a linear function near $x$. That is

$$\lim_{h \to 0} \frac{f(x+h)-hf'(x)}{h} = 0$$

and $f'(x)$ is the only such number. You might try thinking about this with some examples: for instance $f(x)=x^2+x,f'(x)=2x+1,f'(1)=3$. How does $\frac{(1+h)^2+(1+h)-3h}{h}$ behave near $h=0$?

Ian
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Visual intuition isn't reliable, particularly when one deals with multidimensional functions and even with single-variable functions whose graphs are not necessarily trivial. In any case, visual reasoning is likely to involve unjustified generalizations.

One robust method of conceptualizing derivatives is to regard the derivative of $f$ at $x_0$ as characterizing the function $f$ at $x_0$ (not completely, but at least partially).

By the definition of the derivative of a single variable function, we have: link to definition of the derivative.

Therefore, from this inequality, there are myriad conclusions that we can make regarding the values that $f$ can take across various vicinities of $x_0$.

Higher degree derivatives provide additional characterization of $f$.

If you don't understand what I am referring to by characterization, consider that the quadratic function $f(x) = ax^2 + bx + c$ is characterized by (at least partially) by $b^2 - 4ac$, since the value of this formula determines how many real roots $f(x)$ can have (and knowing this gives us much insight to the general behaviour of $f(x)$).