Trigonometric Derivation

Today I have a question about differentiation with the definition of differentiation, which means using the lim_{h -> 0} (f(x + h) - f(x)) / h. However, today it is going to be about trigonometric functions (it's always about trigonometry, isn't it?).

Differentiate cos(x) using the definition of differentiation.

First, the most important limit to remember is this: lim_{x -> 0} sin(x) / x = 1. When you have this in mind, you can solve anything (well, many things).

Let's write what we need to find:
lim_{h -> 0} (cos(x + h) - cos(x)) / h

Using the identity of the cosine of sums of angles, we can get:
lim_{h -> 0} (cos(x)cos(h) - sin(x)sin(h) - cos(x)) / h

Now follow closely:
lim_{h -> 0} ((cos(x)cos(h) - cos(x)) / h - sin(x)sin(h) / h)
lim_{h -> 0} (cos(x)cos(h) - cos(x)) / h - lim_{h -> 0} sin(x)sin(h) / h
lim_{h -> 0} cos(x)(cos(h) - 1) / h - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(cos(h) - 1)(cos(h) + 1)/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(cos²(h) - 1)/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h
lim_{h -> 0} cos(x)(-sin²(h))/h(cos(h) + 1) - sin(x) * lim_{h -> 0} sin(h) / h

Now watch the beauty of mathematics reveal itself:
L'Hopitals rule says that to find lim_{x -> a} f(x)/g(x), you can also find lim_{x -> a} f'(x)/g'(x). Let's do that on the first limit:
lim_{h -> 0} -sin²(h) / h
= lim_{h -> 0} (-sin²(h))' / h'
= lim_{h -> 0} -2sin(h)cos(h) / 1
As h approaches 0, sin(h) also approaches zero, so the entire limit is zero, and thus the whole first part of the first big limit.

This leaves us with:
- sin(x) * lim_{h -> 0} sin(h) / h
As you can see, we have lim_{h -> 0} sin(h) / h = 1
That means:
-sin(x) * lim_{h -> 0} sin(h) / h = -sin(x) * 1 = -sin(x)

So the entire derivative equals -sin(x).

That's why super math tips exists. Or you can simply use Wikipedia's list of trigonometric derivatives (but where is the fun in that?).

Nadav

nadavs