What about the derivative of the sine function? The rules for derivatives that we have are no help, since $\sin x$ is not an algebraic function. We need to return to the definition of the derivative, set up a limit, and try to compute it. Here's the definition: $${d\over dx}\sin x = \lim_{\Delta x\to0} {\sin(x+\Delta x)-\sin x \over \Delta x}.$$ Using some trigonometric identities, we can make a little progress on the quotient: \eqalign{ {\sin(x+\Delta x)-\sin x \over \Delta x}&={\sin x \cos \Delta x + \sin \Delta x \cos x - \sin x\over \Delta x}\cr &=\sin x{\cos \Delta x - 1\over \Delta x}+\cos x{\sin\Delta x\over \Delta x}.\cr }

This isolates the difficult bits in the two limits $$\lim_{\Delta x\to0}{\cos \Delta x - 1\over \Delta x}\quad\hbox{and}\quad \lim_{\Delta x\to0} {\sin\Delta x\over \Delta x}.$$ Here we get a little lucky: it turns out that once we know the second limit the first is quite easy. The second is quite tricky, however. Indeed, it is the hardest limit we will actually compute, and we devote a section to it.