Question

Differentiate each of the following from first principles :

$\left(i\right)ta{n}^{2}x$

$\left(ii\right)tan(2x+1)$

$\left(iii\right)tan2x$

$\left(iiv\right)\sqrt{tanx}$

Answer 1

(i) We have,

$f\left(x\right)=ta{n}^{2}x$

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

$=li{m}_{h\to 0}\frac{ta{n}^2(x+h)-ta{n}^{2}x}{h}$

$=li{m}_{h\to 0}\frac{\left\{tan\right(x+h)+tanx\}\left\{tan\right(x+h)-tanx\}}{h}$

$[\because ta{n}^{2}A-ta{n}^{2}B=(tanA+tanB\left)\right(tanA-tanB\left)\right]$

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

$li{m}_{h\to 0}\frac{sin(2x+h)}{h.cos(x+h)cosx}\times \frac{sinh}{cos(x+h)cosx}$

$=li{m}_{h\to 0}\frac{sinh}{h}\times \frac{sin2x}{co{s}^{2}x.co{s}^2(x+h)}$

$=li{m}_{h\to 0}\frac{sin2x}{co{s}^{2}x.co{s}^{2}x}[\because li{m}_{h\to 0}\frac{sinh}{h}=1]$

$=li{m}_{h\to 0}\frac{2sinx.cosx}{co{s}^{2}x}\times \frac{1}{co{s}^{2}x}[sin2x=2sinxcosx]$

$=2tanx.se{c}^{2}x$

(ii) We have,

$f\left(x\right)=tan(2x+1)$

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

$=li{m}_{h\to 0}\frac{tan2(x+h)+tan(2x+1)}{h}$'

$=li{m}_{h\to 0}\frac{sin(2x+2h+1-2x-1)}{h.cos\left\{2\right(x+h)+1\}cos(2x+1)}[\because tanA-tanB=\frac{sin(A-B)}{cosA.cosB}]$

$=li{m}_{h\to 0}\frac{2.sin2h}{2h.cos\{2x+2h+1\}cos(2x+1)}$

Multiplying both, Numerator and Denominator by 2.

$\therefore li{m}_{h\to 0}\left(\frac{sin2h}{2}\right)\times \frac{1}{cos\{2x+2h+1\}cos(2x+1)}$

$=\frac{2}{co{s}^2(2x+1)}[\because \frac{sin2h}{h}=1]$

$=2se{c}^2(2x+1)[\because se{c}^{2}x=\frac{1}{co{s}^{2}x}]$

$=2se{c}^2(2x+1)$

(iii) We have,

$f\left(x\right)=tan2x$

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

$=li{m}_{h\to 0}\frac{tan2(x+h)-tan2x}{h}=li{m}_{h\to 0}\frac{sin(2x+2h-2x)}{h.cos(2x+2h)cos2x}[\because tanA-tanB=\frac{sin(A-B)}{cosA.cosB}]$

$=li{m}_{h\to 0}\frac{sin2h}{h.cos(2x+2h)cos2x}$

$=li{m}_{h\to 0}\left(\frac{sin2h}{2h}\right)\times \frac{1\times 2}{cos(2h+2x)cos2x}$

$=\frac{2}{cos2x.cos2x}[\because li{m}_{h\to 0}\frac{sinh}{h}=1]$

$=2se{c}^{2}2x[\because \frac{1}{co{s}^{2}x}=se{c}^{2}x]$

(iv) We have,

$f\left(x\right)=\sqrt{tanx}$

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

$=li{m}_{h\to 0}\frac{\sqrt{tan(x+h)}-\sqrt{tanx}}{h}$

Multplying Numerator and Denominator by $\sqrt{tan(x+h)}+\sqrt{tanx}$

$=li{m}_{h\to 0}\frac{tan(x+h)-tanx}{(\sqrt{tan(x+h)}+\sqrt{tanx})}=li{m}_{h\to 0}\frac{sin(x+h-x)}{h.cos(x+h)cosx(\sqrt{tan(x+h)}+\sqrt{tanx})}$

$=li{m}_{h\to 0}\frac{sinh}{h}\times \frac{1}{cos(x+h)cosx(\sqrt{tan(x+h)}+\sqrt{tanx})}$

$=li{m}_{h\to 0}\frac{1}{co{s}^2.2\sqrt{tanx}}[\because li{m}_{h\to 0}\frac{sinh}{h}=1]$

$=\frac{1}{2}\frac{se{c}^{2}x}{\sqrt{tanx}}[\because \frac{1}{co{s}^{2}x}=se{c}^{2}x]$