Question

differentiate x/sinx using first principle ​

Answer 1

Step-by-step explanation:

$y = f(x) = \frac{x}{ \sin(x) }$

$\frac{dy}{dx} = \lim_{h \rarr0} \frac{f(x + h) - f(x)}{h}$

$\frac{dy}{dx} = \lim_{h \rarr0} \frac{ \frac{x + h}{ \sin(x + h) } - \frac{x}{ \sin(x) } }{h}$

$\frac{dy}{dx} = \lim_{h \rarr0} \frac{x( \sin(x) - \sin(x + h)) - h \sin(x) }{h \sin(x + h) \sin(x) }$

$\frac{dy}{dx} = \lim_{h \rarr0} \frac{x( \sin(x) - \sin(x) \cos(h) - \cos(x) - \sin(h)) }{h \sin(x + h) \sin(x) } - \lim_{h \rarr0} \frac{1}{ \sin(x + h) }$

$= \lim_{h \rarr0} \frac{x \sin(x)(1 - \cos(h) ) - x \cos(x) \sin(h) }{h \sin(x + h) \sin(x) } - \frac{1}{ \sin(x) }$

$= \lim_{h \rarr0} \frac{x(1 - \cos(h) )}{h \sin(x + h) } - \lim_{h \rarr0} \frac{ \cos(x) \sin(h) }{h \sin(x + h) } - \cosec(x)$

$= \lim_{h \rarr0} \frac{1 - \cos(h) }{h} .\lim_{h \rarr0} \frac{x}{ \sin(x + h) } - \lim_{h \rarr0} \frac{ \sin(h) }{h} .\lim_{h \rarr0} \frac{ \cot(x) }{ \sin(x + h) } - \cosec(x)$

$= 0 \times \frac{x}{ \sin(x) } - 1 \times \frac{ \cos(x) }{ \sin ^{2} (x) } - \frac{1}{ \sin(x) }$

$= - \frac{( \cos(x) + \sin(x)) }{ \sin^{2} (x) }$