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.

This Series
I – Introduction
II – Trigonometry
III – Differentiation Rules


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}


\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.


  • 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.


f = u^av^bw^c

is a function mapping reals to reals.


\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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