Differentiation

Maths Help suggests:

First of all, it really is true that the derivative of sin(x) is cos(x).
We don't agree that this result is "strange" - it is actually very neat!

A formal proof needs quite a lot of background maths - perhaps that is
why your lecturer told you not to worry about it for now. But it is
quite a useful piece of theory, so we will outline it here . . .

First, remember that for any function, the derivative dy/dx represents the
gradient at a particular point P with coordinates P(x,y).
Look at the diagram below:

The gradient of the curve y = f(x) at point P is the same as the gradient
of the tangent to the curve at point P (shown in red).
This is approximately the same as the gradient of the line joining
P and a nearby point Q (shown in blue).
h stands for the small distance along the x axis that Q is away from P.
Do you agree that the closer Q gets to P, the closer the line PQ gets to
the red tangent at P?

Therefore, we can say that the gradient at P (red) is essentially the same as
the gradient of PQ (blue) as the distance h tends to zero.

Now the curve you are interested in is y = sin(x).
So P has coordinates (x , sin(x)) and Q has coordinates (x+h , sin(x+h)).
The gradient of PQ (look at the blue triangle) is "y-step over x-step"

We now need to simplify this, using an important result in trigonometry:

• sin(A + B) = sin(A).cos(B) + cos(A).sin(B)

It is now time to use two more important trig results:

• If h tends to 0 (which we are assuming, because Q gets close to P)
  • cos(h) = 1
  • sin(h) = h

Therefore the gradient of the tangent at P is given by cos(x),
which is the same as saying that the derivative of sin(x) is cos(x).