# 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!

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

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

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

$\displaystyle=\cos(x)$ |

## 10.11.2 Derivative of inverse trig functions

## 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),

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,

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),

## 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).

We start with the identity

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

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

## 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

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

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.