Question

$\frac{d\left(\sin x\right)}{dx}.$

• A. $\cos x$
• B. $\sec x$
• C. $-\cos x$
• D. $-\tan x$

Answer 1

Answer: (A)

$y=\sin x=f\left(x\right),$(say)
$y=\delta y=\sin (x+\delta x)=f(x+\delta x)$
$\therefore \delta y=(x+\delta x)-\sin x=f(x+\delta x)-f\left(x\right)$
$\therefore \frac{\delta y}{\delta x}=\frac{\sin (x+\delta x)-\sin x}{\delta x}=\frac{f(x+\delta x)-f\left(x\right)}{\delta x}$
$\therefore \frac{\delta y}{\delta x}=\underset{\delta x\to 0}{lim}\frac{\sin (x+\delta x)-\sin x}{\delta x}$
$=li{m}_{\delta x\to 0}=\frac{f(x+\delta x)-f\left(x\right)}{\delta x}$
$=\underset{\delta x\to 0}{lim}=\frac{2\cos (x+\frac{1}{2}\delta x)\sin \frac{1}{2}\delta x}{\delta x}$
$=\underset{\delta x\to 0}{lim}\cos (x+\frac{\delta x}{2})\frac{\sin \left(\delta x/2\right)}{\left(\delta x/2\right)}$
$=\cos x.1=\cos x(\because \underset{\delta x\to 0}{lim}\frac{\sin \theta }{\theta }=1.)$