Derivative of sec(x) using First Principle of Derivative

We will use the first principle of derivatives to prove that the derivative of

\sec(x)

\tan(x)\sec(x)

, or in other words, by the definition of the derivative.

Proof. Let

f(x) = \sec(x) = \frac{1}{\cos(x)}

. Then

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

We will use the following identity

\begin{align*} \cos(A) - \cos(B) = -2\sin(\frac{A + B}{2})\sin(\frac{A - B}{2}) \end{align*}

In our case, that would mean

\begin{align*} \cos(x) - \cos(x + h) = -2\sin(\frac{2x + h}{2})\sin(\frac{-h}{2}) \end{align*}

Further, we do from this article that

\lim_{h \rightarrow 0} \frac{\sin(h/2)}{h/2} = 1

and

\sin(-h/2) = -\sin(h/2)

. This results

\begin{align*} \lim_{h \rightarrow 0} \frac{\frac{-2\sin(\frac{2x + h}{2})\sin(\frac{-h}{2})}{\cos(x + h)\cos(x)}}{h} &= \lim_{h \rightarrow 0} -\frac{\sin(-h/2)}{h/2} \cdot \lim_{h \rightarrow 0} \frac{\sin(\frac{2x + h}{2})}{\cos(x + h)\cos(x)} \\ &= \lim_{h \rightarrow 0} \frac{\sin(h/2)}{h/2} \cdot \lim_{h \rightarrow 0} \frac{\sin(\frac{2x + h}{2})}{\cos(x + h)\cos(x)} \\ &= \lim_{h \rightarrow 0} \frac{\sin(\frac{2x + h}{2})}{\cos(x + h)\cos(x)} \\ &= \frac{\sin(x)}{\cos^2(x)} \\ &= \frac{\tan(x)}{\cos(x)} \\ &= \tan(x)\sec(x). \end{align*}

So this concludes that

f'(x) = \tan(x)\sec(x)

Derivative of sec(x) using First Principle of Derivatives