Question

From the first principal find the derivative of 1/ sinx

Useless answers will be reported​

Answer 1

hope it helps .................

Answer 2

ANSWER:

• d/dx( cosec x) = - Cot x cosec x

GIVEN:

• 1 / sinx

TO FIND:

• Derivative of 1 / sinx using first principle.

FORMULAE:

$\displaystyle \frac{df(x)}{dx} = \lim _{h\to0} \frac{f(x + h) - f(x)}{h}$

EXPLANATION:

${ \displaystyle \dfrac{d \left (\dfrac{1}{sin \ x} \right )}{dx} = \lim _{h\to0} \frac{ \dfrac{1}{sin(x + h) }- \dfrac{1}{sin x} }{h} }$

${ \displaystyle \dfrac{d \left (cosec \ x \right )}{dx} = \lim _{h\to0} \frac{ \dfrac{sin x - sin(x + h)}{sin \ x \ sin(x + h)} }{h}}$

${ \displaystyle \dfrac{d \left (cosec \ x \right )}{dx} = \lim _{h\to0} \dfrac{sin x - sin(x + h)}{h(sin \ x \ sin(x + h))} }$

★ Sin A - Sin B = 2Cos[(A + B)/2] × Sin[(A - B)/2]

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

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

★ Sin (- x) = - Sin x

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

$= \lim _{h\to0} \dfrac{ - cos(x + \frac{ h}{2} )sin( \frac{h}{2} )}{ \dfrac{h}{2} (sin \ x \ sin(x + h))}$

$\bold{\bigstar{\gray{\frac{ \large{sin} \left( \dfrac{h}{2} \right) }{ \dfrac{h}{2} } = 1}}}$

$= \lim _{h\to0} \dfrac{ - cos(x + \frac{ h}{2} )}{ (sin \ x \ sin(x + h))}$

$= - \dfrac{cos(x + \frac{ 0}{2} )}{ sin \ x \ sin(x + 0)}$

$= - \dfrac{cos \ x}{ sin \ x \ sin \ x}$

★ cos x / sin x = cot x

★ 1 / sin x = cosec x

$= - cosec \ x \ cot \ x$

$\dfrac{d \left (\dfrac{1}{sin \ x} \right )}{dx} = - cosec \ x \ cot \ x$