Differentiation rules have been bugging me ever since I took Basic Calculus. I believed I"d construct some intuitive knowledge of them inevitably, but so far all my various other steustatiushistory.org courses (including Multivariable Calculus) take the rules for granted.

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I understand exactly how to prove some of the rules. The problem is that algebra manipulation alone isn"t quite convincing to me. Is tbelow any kind of possibility of expertise why the algebra happens to job-related that way? For instance, why do the slopes of the tangent line to the parabola x^2 take place to be determined by 2x? Looking at it graphically, there"s no means I could"ve told that.

Any sources extending this concern (books; internet sites; etc) would be incredibly considerably appreciated. Thanks in advance.

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edited Mar 25 "16 at 13:51

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The crucial intuition, first of all, is that the product of 2 tiny differences is negligible. You can intuit this just by doing computations:

$$3.000001 cdot 2.0001 = 6.0003020001$$

If we are doing any kind of type of rounding of hand computations, we"d likely round away that $0.0000000001$ part. If you were doing computations to eight considerable digits, a value $v$ is really a worth in a variety approximately of $vleft(1 pm 10^-8 ight)$ and the error once you multiply $v_1$ by $v_2$ is practically completely $10^-8|v_1v_2|$. The other component of the error is so tiny you"d most likely neglect it.

Case: $f(x)=x^2$

Now, take into consideration a square with corners $(0,0), (0,x), (x,0), (x,x)$. Grow $x$ a little little, and also you watch the area grows by proportionally by the size of 2 of the edges, plus a tiny little square. That tiny square is negligible.

This is a tiny harder to visualize for $x^n$, yet it actually works the very same method when $n$ is a positive integer, by considering an $n$-dimensional hypercube.

This geometric factor is also why the circumference of a circle is equal to the derivative of its location – if you rise the radius a small, the area is raised by about that "little" times the circumference. So the derivative of $pi r^2$ is the circumference of the circle, $2pi r$.

It"s also a means to understand also the product rule. (Or, indeed, FOIL.)

Case: The chain rule

The chain ascendancy is much better viewed by considering an odd-shaped tub. Let"s say that when the volume of the water in a tube is $v$ then the tub is filbrought about depth $h(v)$. Then assume that we have actually a hose that, between time $0$ and time $t$, has actually sent a volume of $v(t)$ water.

At time $t$, what is the rate that the elevation of the water is increasing?

Well, we know that as soon as the present volume is $v$, then the rate at which the elevation is increasing is $h"(v)$ times the price the volume is raising. And the price the volume is enhancing is $v"(t)$. So the price the height is raising is $h"(v(t)) cdot v"(t)$.

Case: Inverse function

This is the one instance wbelow it is evident from the graph. When you flip the works with of a Cartesian airplane, a line of slope $m$ gets sent to a line of slope $1/m$. So if $f$ and $g$ are inverse functions, then the slope of $f$ at $(x,f(x))$ is the inverse of the slope of $g$ at $(f(x),x)=(f(x),g(f(x)))$. So $g"(f(x))=1/f"(x)$.

$x^2$ revisited

Another way of managing $f(x)=x^2$ is reasoning aobtain of location, yet reasoning of it in terms of units. If we have actually a square that is $x$ centimeters, and also we change that by a little amount, $Delta x$ centimeters, then the area is $x^2steustatiushistory.orgrmcm^2$ and also it goes to approximately $f(x+Delta x)-f(x)=f"(x)Delta x$.

On the various other hand, if we measure the square in meters, it has side length $x/100$ meters and also area $(x/100)^2$. The change in the side size is $(Delta x)/100$ meters. So the intended location readjust is $f"(x/100)cdot (Delta x)/100$ square meters. But this difference should be the very same, so$$f"(x)Delta x = f"(x/100)cdotfracDelta x100cdot left(100^2 extm^2/ extcm^2 ight) = 100 f"(x/100)$$

More primarily, then, we check out that $f"(ax)=af"(x)$ as soon as $f(x)=x^2$ by changing devices from centimeters to a unit that is $1/a$ centimeters.

So we see that $f"(x)$ is linear, although it does not define why $f"(1)=2$.

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If you do the same for $f(x)=x^n$, with devices $mu$ and also another unit $ ho$ where $a ho = mu$, then you acquire that the a readjust in volume once transforming by $Delta x,mu$ is $f"(x)Delta x,mu^n$. It is also $f"(ax)cdot a(Delta x), ho^n$. Because $mu/ ho = a$, this indicates $f"(ax) =a^n-1f"(x)$.