## Visualizing Elementary Calculus: Differentiation Rules 1

The basic rules of differentiation are linearity, the product rule, and the chain rule. Once we start graphing functions, we’ll revisit these rules.

### Linearity

The linearity of differentials means

$\textrm{d}(\alpha u + \beta v) = \alpha \textrm{d}u + \beta \textrm{d}v$

$\alpha$ and $\beta$ are constants, while $u$ and $v$ might change.

This looks obvious, but here’s a quick sketch.

First we’ll look at $\textrm{d}(\alpha u)$. Construct a right triangle with base 1 and hypotenuse $\alpha$. Then extend the base by length $u$. This creates a larger, similar triangle. The hypotenuse must be $\alpha$ times the base, so the hypotenuse is extended by $\alpha u$.

Then increase $u$ by $\textrm{d}u$. This induces an increase $\textrm{d}(\alpha u)$ in the hypotenuse.

We draw an original blue triangle with base 1 and hypotenuse alpha. Then it's extended to the dark green triangle, adding u to the base and alpha*u to the hypotenuse. Finally, we increment u by du and observe the effect.

The little right triangle made by $\textrm{d}u$ and $\textrm{d}(\alpha u)$ is similar to the original, so

$\frac{\textrm{d}(\alpha u)}{\textrm{d}u} = \frac{\alpha}{1}$

or

$\textrm{d}(\alpha u) = \alpha \textrm{d}u$

Next look at $\textrm{d}(u + v)$. $u+v$ is just two line segments laid one after the other. We increase the lengths by $\textrm{d}u$ and $\textrm{d}v$ and see what the change in the total length $\textrm{d}(u+v)$ is.

The total change is equal to the sum of the changes.

$\textrm{d}(u + v) = \textrm{d}u + \textrm{d}v$

These rules combine to give the rule for linearity

$\textrm{d}(\alpha u + \beta v) = \alpha \textrm{d}u + \beta \textrm{d}v$

### The Product Rule

The product rule is

$\textrm{d}(uv) = u\textrm{d}v + v\textrm{d}u$

To show this, we need a line segment with length $uv$.

Start by drawing $u$, then drawing a segment of length 1 starting at the same place as $u$ and going an arbitrary direction.

Close the triangle. Extend the segment of length 1 by $v$, and close the new triangle. We’ve now extended the base by $uv$.

Construction of length u*v, by similar triangles.

Increase $u$ by $\textrm{d}u$ and $v$ by $\textrm{d}v$. This results in several changes to $uv$.

The segment $uv$ has a little bit chopped off on the left, since $\textrm{d}u$ cuts into the place where it used to be.

$uv$ is also extended twice on the right. The first extension is the projection of $\textrm{d}v$ down onto the base. All such projections multiply the length by $u$, so the piece added is $u\textrm{d}v$.

Finally there is a piece added from the very skinny tall triangle. It is similar to the skinny, short triangle created by adding $\textrm{d}u$ to $u$. The tall triangle is $(1+v)$ times as far from the bottom left corner as the short one, so it is $(1+v)$ times as big. Since the base of the short one is $\textrm{d}u$, the base of the tall one is $(1+v)\textrm{d}u$.

Combining all three changes to $uv$, one subtracting from the left and two adding to the right, we get

$\textrm{d}(uv) = -\textrm{d}u + u\textrm{d}v + (1+v)\textrm{d}u = u\textrm{d}v + v\textrm{d}u$

This is the product rule. We’ll give another visual proof in the exercises.

### The Chain Rule

Suppose we want $\textrm{d}\sin x^2$. (There’s no particular reason I can think of to want that, but we have a limited milieu of functions at hand right now.)

We know $\textrm{d}(\sin\theta) = \cos\theta\textrm{d}{\theta}$. Let $\theta = x^2$.

$\textrm{d}(\sin x^2) = \cos(x^2)\textrm{d}(x^2)$

But we already know that $\textrm{d}(x^2) = 2x\textrm{d}x$, so substitute that in to get

$\textrm{d}(\sin x^2) = \cos(x^2)2x\textrm{d}x$

This is called the chain rule. A symbolic way to right it is

$\frac{\textrm{d}f}{\textrm{d}t} = \frac{\textrm{d}f}{\textrm{d}x}\frac{\textrm{d}x}{\textrm{d}t}$

Suppose you are hiking up a mountain trail. $f$ is your height above sea level. $x$ is the distance you’ve gone down the trail. $t$ is the time you’ve been hiking.

$\textrm{d}f/\textrm{d}t$ is the rate you are gaining height. According to the chain rule, you can calculate this rate by multiplying the slope of the trail $\textrm{d}f/\textrm{d}x$ to your speed $\textrm{d}x/\textrm{d}t$.

### Exercises

• Show that the linearity rule $\textrm{d}(\alpha u) = \alpha \textrm{d}u$ is a special case of the product rule.
• What is the derivative of $A\sin\theta + C\cos\theta$ with respect to $\theta$? Take the derivative with respect to $\theta$ of that. (This is called a “second derivative”.) What do you get? (Answer: -1 times the original function)
• Use the product rule to prove by induction that the derivative of $x^n$ is $n x^{n-1}$ for all positive integers $n$.
• Apply the product rule to $x^nx^{-n} = 1$ to prove that the “power rule” from the previous question holds for all integers $n$.
• Look back at the arguments from the introduction. Draw a rectangle with one side length $u$ and one side length $v$. Its area is $uv$. Use this to prove the product rule.
• Apply the chain rule to $(x^{1/n})^{n} = x$ to find the derivative of $x^{1/n}$ with respect to $x$ for all integers $n$ (Answer: $\frac{1}{n} x^{1/n -1}$)
• Argue that the derivative of $x^{p/q} = \frac{p}{q}x^{p/q - 1}$ for all rational numbers $p/q$.
• Show that the derivative of a polynomial is always another polynomial. Is there any polynomial that is its own derivative? (Answer: no, except zero)
• Combine the product rule with the chain rule to prove the quotient rule $\textrm{d}\frac{u}{v} = \frac{u\textrm{d}v - v\textrm{d}u}{v^2}$

### One Response to “Visualizing Elementary Calculus: Differentiation Rules 1”

1. Mohamed c/raxmaan jama' Says:

First, i greated all muslims by lhe greating of islam which says: “salamu calaykum”. I admire how elementary calculus works , lhe one who discovered or developed it is newton , newton was very lhinker person . If my voice goes to a scientist, it goes to newton! For his hard work on science.