Derivative of tan(x) using First Principle of Derivatives

The Mathematician — Mon, 05 Sep 2022 15:00:46 +0000

Using the first principle of derivatives, we will prove the derivative of a tangent, or in other words, that the derivative

\tan(x)

1/\cos^2(x)

. Proof. Let

f(x) = \tan(x) = \frac{\sin(x)}{\cos(x)}

. Then

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

We will use the next identity:

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

This would imply for us that

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

Therefore we get

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

We do have know that

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

from this article. So continuing where we left:

\begin{align*} \lim_{h \rightarrow 0} \frac{1}{\cos(x + h)\cos(x)} = \frac{1}{\cos(x)\cos(x)} = \frac{1}{\cos^2(x)} = \sec^2(x) \end{align*}

Therefore, we get

f'(x) = \frac{1}{\cos^2(x)} = \sec^2(x)

derivative of tan x Archives - Epsilonify