Derivative of csc(x) using First Principle of Derivatives

The Mathematician — Wed, 14 Sep 2022 13:00:24 +0000

Using the first principle of derivatives, we will show that the derivative of

\csc(x)

is equal to

-\csc(x)\cot(x)

. In other words,

\frac{d}{dx}\cot(x) = -\csc(x)\cot(x)

. The

\csc

is also called the cosecant.

Proof. Let

f(x) = \csc(x) = \frac{1}{\sin(x)}

. Then

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

We will use the following identity

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

This results

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

Therefore, we get the following equality

\begin{align*} \lim_{h \rightarrow 0} \frac{\frac{2\cos(\frac{2x + h}{2})\sin(\frac{-h}{2})}{\sin(x + h)\sin(x)}}{h} &= \lim_{h \rightarrow 0} \frac{\sin(-h/2)}{h/2} \cdot \lim_{h \rightarrow 0} \frac{\cos(\frac{2x + h}{2})}{\sin(x + h)\sin(x)} \\ &= -1 \cdot \lim_{h \rightarrow 0} \frac{\cos(\frac{2x + h}{2})}{\sin(x + h)\sin(x)} \\ &= -1 \cdot \frac{\cos(x)}{\sin^2(x)} \\ &= - \frac{1}{\sin(x)\tan(x)} \\ &= - \csc(x)\cot(x) \end{align*}

where we have seen that

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

\frac{1}{\sin(x)} = \csc(x)

and

\frac{1}{\tan(x)} = \cot(x)

. So we have that

f'(x) = -\csc(x)\cot(x)

Derivative of csc x using First Principle Method Archives - Epsilonify