# How do you differentiate f(x)=sin(x) from first principles?

Jun 12, 2018

$\frac{d}{\mathrm{dx}} \sin x = \cos x$

#### Explanation:

By definition of the derivative:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

So with $f \left(x\right) = \sin x$ we have;

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\sin \left(x + h\right) - \sin x}{h}$

Using $\sin \left(A + B\right) = \sin A \cos B + \sin B \cos A$ we get

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{\sin x \cos h + \sin h \cos x - \sin x}{h}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{h \rightarrow 0} \frac{\sin x \left(\cos h - 1\right) + \sin h \cos x}{h}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{h \rightarrow 0} \left(\frac{\sin x \left(\cos h - 1\right)}{h} + \frac{\sin h \cos x}{h}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\lim}_{h \rightarrow 0} \frac{\sin x \left(\cos h - 1\right)}{h} + {\lim}_{h \rightarrow 0} \frac{\sin h \cos x}{h}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(\sin x\right) {\lim}_{h \rightarrow 0} \frac{\cos h - 1}{h} + \left(\cos x\right) {\lim}_{h \rightarrow 0} \frac{\sin h}{h}$

We know have to rely on some standard limits:

${\lim}_{h \rightarrow 0} \sin \frac{h}{h} = 1$, and ${\lim}_{h \rightarrow 0} \frac{\cos h - 1}{h} = 0$

And so using these we have:

$f ' \left(x\right) = 0 + \left(\cos x\right) \left(1\right) = \cos x$

Hence,

$\frac{d}{\mathrm{dx}} \sin x = \cos x$