## Posts Tagged ‘differentiation’

### Visualizing Elementary Calculus: Differentiation Rules 1

March 27, 2011

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 $\textrm{d}\frac{u}{v} = \frac{u\textrm{d}v - v\textrm{d}u}{v^2}$

### Feynman’s Differentiation Trick

May 10, 2009

Feynman’s Tips on Physics describes the following trick to take derivatives quickly.

Suppose

$f = u^av^bw^c$

is a function mapping reals to reals.

Then

$\frac{df}{dx} = a\frac{du}{dx}u^{a-1}v^bw^c + u^ab\frac{dv}{dx}v^{b-1}w^c + u^av^bc\frac{dw}{dx}w^{c-1}$

Next notice that each of these terms almost has $f$ itself in it, so we factor $f$ out.

$\frac{df}{dx} = f\left(a\frac{du/dx}{u} + b\frac{dv/dx}{v} + c\frac{dw/dx}{w}\right)$.

That’s the trick. It’s not incredibly powerful, but here’s how it works on an example:

$f=\frac{\sin^3(x)\sqrt{e^{x+x^2}}}{\left(x^2+1\right)^{1/3}}$.

We’ll let

$u = \sin(x), a=3$
$v = e^{x+x^2}, b=1/2$
$w = x^2 + 1, c=-1/3$

Next we write $f$ over again.

$\frac{df}{dx} = \frac{\sin^3(x)\sqrt{e^{x+x^2}}}{\left(x^2+1\right)^{1/3}}*[\ldots]$.

Then insert $a\frac{du/dx}{u}$.

$\frac{df}{dx} = \frac{\sin^3(x)\sqrt{e^{x+x^2}}}{\left(x^2+1\right)^{1/3}}*\left[3\frac{\cos(x)}{\sin(x)}+\ldots\right]$

And then add $b\frac{dv/dx}{v}$.

$\frac{df}{dx} = \frac{\sin^3(x)\sqrt{e^{x+x^2}}}{\left(x^2+1\right)^{1/3}}*\left[3\frac{\cos(x)}{\sin(x)}+\frac{1}{2}\frac{e^{x+x^2}(1+2x)}{e^{x+x^2}}+\ldots\right]$.

Finish it up with $c\frac{dw/dx}{w}$.

$\frac{df}{dx} = \frac{\sin^3(x)\sqrt{e^{x+x^2}}}{\left(x^2+1\right)^{1/3}}*\left[3\frac{\cos(x)}{\sin(x)}+\frac{1}{2}\frac{e^{x+x^2}(1+2x)}{e^{x+x^2}}+\frac{-1}{3}\frac{2x}{x^2+1} \right]$.

### Calculus Pretest Solutions

May 9, 2009

solutions to the pretest

let me know if you find any errors. no solutions for the proofs, though.

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### Calculus Pretest

May 8, 2009

I’m about to start teaching a friend of mine calculus. He took it in school a while ago, so I made this little pretest to see where we’re starting from. If you understand everything on here, you should be able to get a 5 on the AP calculus exam.

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