Question

Differentiate x cos x by first principle.

Or

Evaluate $li{m}_{y\to 0}\frac{(x+y)sec(x+y)-xsecx}{y}$

Answer 1

Let f(x) = x cos x, then f (x + h) = (x + h) cos (x + h)

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

$=li{m}_{h\to 0}\frac{(x+h)cos(x+h)-xcosx}{h}$

$=li{m}_{h\to 0}\frac{xcos(x+h)-xcosx+hcos(x+h)}{h}$

$=li{m}_{h\to 0}\frac{x\left[cos\right(x+h)-cosx]}{h}+li{m}_{h\to 0}\frac{hcos(x+h)}{h}$

$[\because cosC-cosD=-2sin\frac{C+D}{2}.sin\frac{C-D}{2}]$

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

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

$=li{m}_{h\to 0}-xsin(x+\frac{h}{2})\times li{m}_{h\to 0}\frac{sin\frac{h}{2}}{\frac{h}{2}}+li{m}_{h\to 0}cos(x+h)$

$=-xsinx\times 1+cosx[\because li{m}_{h\to 0}\frac{sinh}{h}=1]$

= cos x - x sin x

Or

We have, $li{m}_{y\to 0}\frac{(x+y)sec(x+y)-xsecx}{y}$

$=li{m}_{y\to 0}\frac{\frac{x+y}{cos(x+y)}-\frac{x}{cosx}}{y}$

$=li{m}_{y\to 0}\frac{(x+y)cosx-xcos(x+y)}{ycos(x+y)cosx}$

$=li{m}_{y\to 0}\frac{xcosx-xcos(x+y)+ycosx}{ycos(x+y)cosx}$

$=li{m}_{y\to 0}\frac{x[cosx-cos(x+y\left)\right]}{ycos(x+y)cosx}+li{m}_{y\to 0}\frac{ycosx}{ycos(x+y)cosx}$

$=li{m}_{y\to 0}\frac{2xsin\left(\frac{x+x+y}{2}\right)sin\left(\frac{xx+y-x}{2}\right)}{ycos(x+y)cosx}+li{m}_{y\to 0}\frac{1}{cos(x+y)}$

$[\because cosC-cosD=2sin\left(\frac{C+D}{2}\right)sin\left(\frac{D-C}{2}\right)]$

$=li{m}_{y\to 0}\frac{2xsin(1+\frac{y}{2})sin\frac{y}{2}}{ycos(x+y)cosx}+li{m}_{y\to 0}\frac{1}{cos(x+y)}$

$=li{m}_{y\to 0}\frac{xsin(x+\frac{y}{2})}{cos(x+y)cosx}\times li{m}_{y\to 0}\frac{sin\frac{y}{2}}{\frac{y}{2}}+li{m}_{y\to 0}\frac{1}{cos(x+y)}$

$=\frac{xsinx}{cosx.cosx}\times 1+\frac{1}{cosx}[\because li{m}_{y\to 0}\frac{sinh}{h}=1]$

= x sec x tan x + sec x