Question

why is $\frac{d}{dx} sinx = cosx ?$

Answer 1

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

y = sinx

Also,

y + δy = sin(x + δx)

Hence,

$\tt{\rightarrow\dfrac{\delta y}{\delta x}=\dfrac{sin(x+\delta x)-sinx}{\delta x}}$

Therefore,

$\tt{\rightarrow\dfrac{dy}{dx}=\lim_{\delta x \to 0}\dfrac{\delta y}{\delta x}}$

$\tt{\rightarrow \lim_{\delta x \to 0}\dfrac{sin(x+\delta x)-sinx}{\delta x}}$

${\boxed{\sf\:{sinC-sinD=2cos(\dfrac{C+D}{2})sin(\dfrac{C-D}{2})}}}$

$\tt{\rightarrow \lim_{\delta x \to 0}\dfrac{2cos(x+\delta x/2)sin(\delta x/2)}{\delta x}}$

$\tt{\rightarrow lim_{\delta x \to 0}cos(x+\dfrac{\delta x}{2})\times lim_{\dfrac{\delta x}{2} \to 0}\dfrac{sin(\delta x/2)}{\delta x/2}}$

= cosx × 1

= cosx

Therefore,

$\tt{\rightarrow\dfrac{dy}{dx}=cosx}$

That is,

$\tt{\rightarrow\dfrac{d}{dx}(sinx)=cosx}$