# What is the derivative of -sin(x)?

Dec 23, 2014

The previous answer contains mistakes. Here is the correct derivation.

First of all, the minus sign in front of a function $f \left(x\right) = - \sin \left(x\right)$, when taking a derivative, would change the sign of a derivative of a function $f \left(x\right) = \sin \left(x\right)$ to an opposite. This is an easy theorem in the theory of limits: limit of a constant multiplied by a variable equals to this constant multiplied by a limit of a variable. So, let's find the derivative of $f \left(x\right) = \sin \left(x\right)$ and then multiply it by $- 1$.

We have to start from the following statement about the limit of trigonometric function $f \left(x\right) = \sin \left(x\right)$ as its argument tends to zero:
${\lim}_{h \to 0} \sin \frac{h}{h} = 1$

Proof of this is purely geometrical and is based on a definition of a function $\sin \left(x\right)$. There are many Web resources that contain a proof of this statement, like The Math Page.

Using this, we can calculate a derivative of $f \left(x\right) = \sin \left(x\right)$:
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\sin \left(x + h\right) - \sin \left(x\right)}{h}$
Using representation of a difference of $\sin$ functions as a product of $\sin$ and $\cos$ (see Unizor , Trigonometry - Trig Sum of Angles - Problems 4) ,
$f ' \left(x\right) = {\lim}_{h \to 0} \frac{2 \cdot \sin \left(\frac{h}{2}\right) \cos \left(x + \frac{h}{2}\right)}{h}$
$f ' \left(x\right) = {\lim}_{h \to 0} \sin \frac{\frac{h}{2}}{\frac{h}{2}} \cdot {\lim}_{h \to 0} \cos \left(x + \frac{h}{2}\right)$
$f ' \left(x\right) = 1 \cdot \cos \left(x\right) = \cos \left(x\right)$

Therefore, derivative of $f \left(x\right) = - \sin \left(x\right)$ is $f ' \left(x\right) = - \cos \left(x\right)$.