Question

Find derivative using the principle.
Cosecx

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

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$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

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$\longrightarrow \: \tt \: \tt \:\lim_{y\to 0}\dfrac{siny}{y} = 1$

$\tt \: sinx - siny = 2sin\bigg( \dfrac{x - y}{2} \bigg)cos\bigg( \dfrac{x + y}{2} \bigg)$

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$\tt \longrightarrow \: Let f(x) = cosecx \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt \longrightarrow \: Change x \: to \: x \: + \: h \: \: \: \: \: \: \: \: \: \: \: \: \\ \longrightarrow \: \tt \: f(x + h) = cosec(x + h)$

☆ Using definition of first principle

$\longrightarrow \: \tt \: f'(x) = \tt \:\lim_{h\to 0}\dfrac{f(x + h) - f(x)}{h}$

$\longrightarrow \: \tt \: f'(x) = \tt \:\lim_{h\to 0}\dfrac{cosec(x + h) - cosecx}{h}$

$\longrightarrow \: \tt \: f'(x) = \tt \:\lim_{h\to 0}\dfrac{\dfrac{1 }{sin(x + h)} - \dfrac{1}{sinx} }{h}$

$\longrightarrow \: \tt \: f'(x) = \tt \:\lim_{h\to 0}\dfrac{sinx - sin(x + h)}{sinx \: sin(x + h) \: h}$

$\tt \: f'(x) = \tt \ \: \: \dfrac{1}{ {sin}^{2}x } \lim_{h\to 0}\dfrac{2 \: sin\bigg(\dfrac{x - x - h}{2} \bigg)cos\bigg( \dfrac{x + x + h}{2} \bigg)}{h}$

$\longrightarrow \: \tt \: f'(x) = \tt \:\dfrac{2cosx}{ {sin}^{2} x} \lim_{h\to 0}\dfrac{sin\bigg(\dfrac{ - h}{2} \bigg)}{h}$

$\longrightarrow \: \tt \: f'(x) = \tt \:\dfrac{2cosx}{ {sin}^{2} x} \lim_{h\to 0}\dfrac{sin\bigg(\dfrac{ - h}{2} \bigg)}{ - 2 \times \dfrac{ ( - h)}{2} }$

$\longrightarrow \: \tt \: f'(x) = - \dfrac{cosx}{ {sin}^{2}x }$

$\longrightarrow \: \tt \: f'(x) = - cosecx \: cotx$

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