10.11 Derivatives of trigonometric functions

What is the derivative of $\sin(x)$? First, we can use the definition of the limit and a little algebra.

We want to rewrite this in terms of the two limits we found in the previous section!

This requires a \saysmart idea here, which is to relate the derivative of $\arctan(x)$, which we don’t know, to the derivative of $\tan(x)$, which we do know. We can start with this formula, which exploits the property that $\arctan$ is the inverse function of $\tan$ (and vice versa),

Then, we can differentiate both sides (using the chain rule), from which we can deduce that

Therefore,

Then, we can use a substitution (another clever idea), in this case by setting $u=\tan(x)$ we then can deduce that

Or, equivalently (as renaming the variable $u$ has no effect on the equation),

The method for finding the derivative of $\arccos(x)$ is pretty much the same as the method for finding the derivative of $\arctan(x)$ (as in Section 10.11.3).

We start with the identity

which we then differentiate. The result of this is that

which we can simplify by computing the parts we know how to compute which allows us to establish the equation

Therefore, we know that

Now, as when differentiating $\arctan(x)$, we can substitute. In this case we will use $u=\cos(x)$ rather than $\tan(x)$, and then write that

This gives us the answer, and if you prefer the variable name $x$ to $u$, then we can just rename it, which means that

This proof/derivation is included purely for completeness and is mechanistically almost entirely the same as the cases of $\arctan$ (Section 10.11.3) and $\arccos$ (Section 10.11.4).

We start with the definition of $\arcsin$, which is that

Differentiating this,

Then, we can substitute $u=\sin(x)$, using which

This is almost what we would like, except that we have a random $\cos(x)$ floating around. This can be removed by noting that $\cos(x)=\sqrt{1-\sin^{2}(x)}$, or $\cos(x)=\sqrt{1-u^{2}}$, i.e.