Question

Find the derivative of the following functions at the indicated points :

$\left(i\right)sinx\text{at}x=\frac{\pi }{2}$

$\left(ii\right)x\text{at}x=1$

$\left(iii\right)2cosx\text{at}x=\frac{\pi }{2}$

$\left(iv\right)sin2x\text{at}x=\frac{\pi }{2}$

Answer 1

(i) We have,

f(x) = sin x

$\because f^{\prime }\left(a\right)=li{m}_{h\to 0}\frac{f(a+h)-f\left(a\right)}{h}$

$=f^{\prime }\left(\frac{\pi }{2}\right)=li{m}_{h\to 0}\frac{f(\frac{\pi }{2}+h)-f\left(\frac{\pi }{2}\right)}{h}$

$=li{m}_{h\to 0}\frac{f(\frac{\pi }{2}+h)-sin\left(\frac{\pi }{2}\right)}{h}$

$=li{m}_{h\to 0}\frac{cosh-1}{h}$

$=li{m}_{h\to 0}\frac{(1-\frac{h^2}{2!}+\frac{h^4}{4!}-.....)-1}{h}$

$[\because cosx=1-\frac{x^2}{2!}+\frac{x^4}{4!}-....]$

$=li{m}_{h\to 0}\frac{(-\frac{h}{2!}+\frac{h^3}{4!}-\frac{h^5}{6!}+......)}{h}$

$=li{m}_{h\to 0}h(-\frac{h}{2!}+\frac{h^3}{4!}-\frac{h^5}{6!}+....)$

$=0$

$\therefore f^{\prime }\left(\frac{\pi }{2}\right)=1$

(ii) We have

$\because f^{\prime }\left(a\right)=li{m}_{h\to 0}\frac{f(a+h)-f\left(a\right)}{h}$

$\because f^{\prime }\left(1\right)=li{m}_{h\to 0}\frac{f(1+h)-f\left(1\right)}{h}$

$=li{m}_{h\to 0}\frac{1+h-1}{h}$

$=li{m}_{h\to 0}1$

$\therefore f^{\prime }\left(1\right)=1$

(iii) We have

$f\left(x\right)=2cosx$

$\because f^{\prime }\left(a\right)=li{m}_{h\to 0}\frac{f(a+h)-f\left(a\right)}{h}$

$=li{m}_{h\to 0}\frac{f(\frac{\pi }{2}+h)-f\left(\frac{\pi }{2}\right)}{h}$

$=li{m}_{h\to 0}\frac{2cos(\frac{\pi }{2}+h)-2cos\left(\frac{\pi }{2}\right)}{h}$

$=li{m}_{h\to 0}\frac{-2sinh-0}{h}$

$=-2[\because li{m}_{\theta \to 0}\frac{sin\theta }{\theta }=1]$

$\therefore f^{\prime }\left(\frac{\pi }{2}\right)=-2$

(iv) We have

$f\left(x\right)=sin2x$

Therefore,

$f^{\prime }\left(a\right)=li{m}_{h\to 0}\frac{f(a+h)-f\left(a\right)}{h}=li{m}_{h\to 0}\frac{f(\frac{\pi }{2}+h)-f\left(\frac{\pi }{2}\right)}{h}$

$=li{m}_{h\to 0}\frac{sin2(\frac{\pi }{2}+h)-sin2\left(\frac{\pi }{2}\right)}{h}$

$=li{m}_{h\to 0}\frac{sin(\frac{\pi }{2}\times 2+2h)-sin\left(\pi \right)}{h}$

$=li{m}_{h\to 0}\frac{-cos2h-0}{h}$

$=-2$

$\text{Therefore}{f}^{\prime }\left(\frac{\pi }{2}\right)=-2$