Using first principle find the derivative of cosec x.

Engineering

Mathematics

Trigonometric Limits

Concept of Derivative

Derivative of Standard Functions

Question

Using first principle find the derivative of cosec x.

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Let y = f(x) = cosec x

∴ f(x + h) = cosec(x + h)

$∴ \frac{dy}{dx} = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

$= lim_{h \to 0} \frac{cosec (x + h) - cosecx}{h}$

$= lim_{h \to 0} \frac{1}{h} [\frac{1}{\sin (x + h)} - \frac{1}{\sin x}]$

$= lim_{h \to 0} \frac{\sin x - \sin (x + h)}{h \cdot \sin x \cdot \sin (x + h)}$

$= lim_{h \to 0} \frac{2 \cos (\frac{2 x + h}{2}) \sin (- \frac{h}{2})}{h \cdot \sin xsin (x + h)}$

$= - lim_{h \to 0} \frac{\cos (x + \frac{h}{2})}{\sin x \cdot \sin (x + h)} \cdot lim_{h \to 0} \frac{\sin \frac{h}{2}}{h / 2}$

$= - \frac{\cos x}{\sin x \cdot \sin x} \cdot lim_{z \to 0} \frac{\sin z}{z}$

$[z = \frac{h}{2} Then, z \to 0 when \Rightarrow h \to 0]$

$= \frac{- \cos x}{\sin x \cdot \sin x} \cdot 1$

$= - \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x}$

= – cosec x, cot x

$∴ \frac{dy}{dx} = - cosecx \cdot \cot x$