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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

Watboy

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asked Oct 2 "15 at 15:58

Matt24Matt24

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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 $f"(ax) =a^n-1f"(x)$.