Question

Differentiation of d/dx ( sinx)

Answer 1

d/DX=sin x= cosx

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Answer 2

Answer:

$\therefore \: \: \: \boxed{ \rm \frac{dy}{dx}sinx = cosx}$

Explanation:

$\large \rm Let \: \: f(x) = sinx$

$\rm\therefore \: f(x+h) = sin(x+h), \: (Putting x= x + h)$

We know that :-

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

$\rm{ \frac{dy}{dx} = \displaystyle \lim_{ \rm h \to 0} \frac{ \rm sin(x + h) - sinx}{ \rm h} }$

$\scriptsize \rm{ = \displaystyle \lim_{ \rm h \to 0} \frac{ \rm 2cos \frac{x + h + x}{2} \: sin \frac{x + h - x}{2} }{ \rm h} } \: [ \because \: sinC-sinD=2cos \frac{C+D}{2} \: sin \frac{C - D}{2} ]$

$\rm{ = \displaystyle \lim_{ \rm h \to 0} \frac{ \rm 2cos( \frac{2x + h}{2} ) sin \frac{h}{2} }{ \rm h} }$

$= \displaystyle \lim_{ \rm h \to 0} \rm cos(x + \frac{h}{2} ). \displaystyle \lim_{ \rm h \to 0} \rm{ \frac{sin \frac{h}{2} }{ \frac{h}{2} } }$

$\rm= cos(x + \frac{0}{2} ) \times 1 \: \: [ \because \: \displaystyle \lim_{ \rm x \to0} \rm\frac{sinx}{x} = 1 ]$

$\rm = cos(x + 0)$

$\rm = cosx$

$\therefore \: \: \: \boxed{ \rm \frac{dy}{dx}sinx = cosx}$