Derivative of sin(x) using First Principle of Derivatives

In this article, we will prove the derivative of sinus, or in other words, the derivative of

\sin(x)

, using first principle of derivatives. We know that the derivative of

\sin(x)

\cos(x)

, but we would also like to see how to prove that by the definition of the derivative.

Proof. Let

f(x) = \sin(x)

. Then

\begin{align*} f'(x) &= \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \rightarrow 0} \frac{\sin(x + h) - \sin(x)}{h} \end{align*}

We do have the following equality:

\begin{equation*} \sin(x + y) = \sin(x)\cos(y) + \cos(x)\sin(y). \end{equation*}

Applying that with

y = h

, we get

\begin{align*} &\lim_{h \rightarrow 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h} = \\ &\lim_{h \rightarrow 0} \frac{\sin(x)(\cos(h) - 1) + \cos(x)\sin(h)}{h} = \\ &\lim_{h \rightarrow 0} \frac{\sin(x)(\cos(h) - 1)}{h} + \lim_{h \rightarrow 0} \frac{\cos(x)\sin(h)}{h} = \\ &\lim_{h \rightarrow 0} \sin(x) \lim_{h \rightarrow 0}\frac{(\cos(h) - 1)}{h} + \lim_{h \rightarrow 0} \cos(x) \lim_{h \rightarrow 0}\frac{\sin(h)}{h} \end{align*}

Now we know from the article that

\lim_{h \rightarrow 0}\frac{\sin(h)}{h} = 1

and

\lim_{h \rightarrow 0}\frac{(\cos(h) - 1)}{h} = 0

from this article. So we get

\begin{equation*} \sin(x) \cdot 0 + \cos(x) \cdot 1 = \cos(x). \end{equation*}

Conclusion: we get the following result

\begin{equation*} f'(x) = \cos(x) \end{equation*}