# 10.11 Derivatives of trigonometric functions

## 10.11.1 Derivative of $\sin(x)$

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

 $\displaystyle\frac{d}{dx}[\sin(x)]$ $\displaystyle=\lim_{h\to 0}\frac{\sin(x+h)-\sin(x)}{h}$ (10.93) $\displaystyle=\lim_{h\to 0}\frac{\sin(x)\cos(h)+\sin(h)\cos(x)-\sin(x)}{h}$ (10.94)

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

 $\displaystyle\lim_{h\to 0}\frac{\sin(x)\cos(h)+\sin(h)\cos(x)-\sin(x)}{h}$ $\displaystyle=\lim_{h\to 0}\left[\frac{\sin(x)(\cos(h)-1)}{h}+\frac{\sin(h)}{h% }\cos(x)\right]$ $\displaystyle=\lim_{h\to 0}\left[-\sin(x)\frac{(1-\cos(h))}{h}+\frac{\sin(h)}{% h}\cos(x)\right]$ $\displaystyle=\lim_{h\to 0}\left[0+1\cos(x)\right]$ $\displaystyle=\cos(x)$

## 10.11.3 Derivative of $\arctan(x)$

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),

$\arctan(\tan(x))=x.$ (10.95)

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

 $\displaystyle\frac{d}{d\left(\tan(x)\right)}\left(\arctan(\tan(x))\right)\frac% {d}{dx}\left(\tan(x)\right)=1$ (10.96) $\displaystyle\implies\frac{d}{d\left(\tan(x)\right)}\left(\arctan(\tan(x))% \right)\left(\sec^{2}(x)\right)=1$ (10.97)

Therefore,

$\frac{d}{d\left(\tan(x)\right)}\left(\arctan(\tan(x))\right)=\frac{1}{\sec^{2}% (x)}$ (10.98)

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

 $\displaystyle\frac{d}{du}(\arctan(u))$ $\displaystyle=\frac{1}{1+\tan^{2}(x)}\text{ by the Pythagorean identity}$ $\displaystyle=\frac{1}{1+u^{2}}$

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

$\frac{d}{dx}(\arctan(x))=\frac{1}{1+x^{2}}$ (10.99)

## 10.11.4 Derivative of $\arccos(x)$

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).

$\arccos(\cos(x))=x,$ (10.100)

which we then differentiate. The result of this is that

 $\displaystyle\left(\frac{d}{d(\cos(x))}\arccos(\cos(x))\right)\frac{d}{dx}(% \cos(x))=\frac{d}{dx}(x),$ (10.101)

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

$\left(\frac{d}{d(\cos(x))}\arccos(\cos(x))\right)(-\sin(x))=1.$ (10.102)

Therefore, we know that

$\left(\frac{d}{d(\cos(x))}\arccos(\cos(x))\right)=-\frac{1}{\sin(x)}.$ (10.103)

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

$\frac{d}{du}\arccos(u)=-\frac{1}{\sqrt{1-u^{2}}}.$ (10.104)

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

$\frac{d}{dx}\arccos(x)=-\frac{1}{\sqrt{1-x^{2}}}.$ (10.105)

## 10.11.5 Derivative of $\arcsin(x)$

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

$\arcsin(\sin(x))=x.$ (10.106)

Differentiating this,

 $\displaystyle\frac{d}{dx}\left(\arcsin(\sin(x))\right)=\frac{d}{dx}(1)$ (10.107) $\displaystyle\iff\frac{d}{d(\sin(x))}(\arcsin(\sin(x)))\frac{d}{dx}\sin(x)=1$ (10.108) $\displaystyle\iff\frac{d}{d(\sin(x))}(\arcsin(\sin(x)))\cos(x)=1$ (10.109) $\displaystyle\iff\frac{d}{d(\sin(x))}(\arcsin(\sin(x)))=\frac{1}{\cos(x)}$ (10.110)

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

$\frac{d}{du}\arcsin(u)=\frac{1}{\cos(x)}.$ (10.111)

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.

$\frac{d}{du}\arcsin(u)=\frac{1}{\sqrt{1-u^{2}}}$ (10.112)