Question

derivative of cosx wrt x
​

Answer 1

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

$: \implies \tt \: cosx - cosy = - 2sin( \frac{x + y}{2} )sin( \frac{x - y}{2})$

$: \implies \tt \: \tt \:\lim_{y\to 0} \dfrac{sin \: y}{y} \: = 1$

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

$: \implies \tt \: Let \: f(x) \: = \: cosx$

$: \implies \tt \: change \: x \: to \: x \: + \: h$

$: \implies \tt \: f(x + h) = cos(x + h)$

★ By using first principal, we get

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

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

$: \implies \tt \: f'(x) =\lim_{h\to 0}\dfrac{ - 2sin( \frac{x + h + x}{2} ) \: sin( \frac{x + h - x}{2}) }{h}$

$: \implies \tt \: f'(x) =\lim_{h\to 0}\dfrac{ - 2sin( \frac{2x + h}{2} ) \: sin( \frac{ h }{2}) }{h}$

$: \implies \tt \: f'(x) = - sinx \times \lim_{h\to 0}\dfrac{sin \frac{h}{2} }{ \frac{h}{2} }$

$: \implies \tt \: f'(x) = - \: sinx$