Question

Differentiate each the following from first principles :

$\left(i\right)xsinx$

$\left(ii\right)x+cosx$

$\left(iii\right)sin(2x-3)$

$\left(iv\right)\sqrt{sin2x}$

$\left(v\right)\frac{sinx}{x}$

$\left(vi\right)\frac{cosx}{x}$

$\left(vii\right)x^{2}sinx$

$\left(viii\right)\sqrt{sin(3x+1)}$

$\left(ix\right)sinx+cosx$

Answer 1

(i) We have,

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

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

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

$=li{m}_{h\to 0}\frac{x\left(sin\right(x+h)-sinx)}{h}+sin(x+h)[sinc-sind=2cos\frac{c+d}{2}sin\frac{c-d}{2}]$

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

$=2x\times cosx\times \frac{1}{2}+sinx=x\times cosx+sinx=sinx+xcosx$

(ii) We have,

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

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

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

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

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

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

(iii) We have,

$f\left(x\right)=sin(2x-3)$

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

$=li{m}_{h\to 0}\frac{sin\left\{2\right(x+h)-3\}-sin(2x-3)}{h}$

$=li{m}_{h\to 0}\frac{2cos\frac{(2x+2h-3+2x-3)}{2}\times \frac{sin(2x+2h-3-2-x+3)}{2}}{h}$

$[\because sinC-sinD=2cos\frac{C+D}{2}sin\frac{C-D}{2}]$

$=li{m}_{h\to 0}2cos(2x-3+h).\frac{sinh}{2}[\because li{m}_{0\to 0}\frac{sin\theta }{\theta }=1]$

$=2cos(2x-3)$

(iv) We have,

$f\left(x\right)=\sqrt{sin2x}$

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

$=li{m}_{h\to 0}\frac{\sqrt{sin(x+h)}-\sqrt{sin2x}}{h}$

Multiplying numerator and Denominator by $\sqrt{sin2(x+h)}+\sqrt{sin2x}$

$=li{m}_{h\to 0}\frac{\sqrt{sin2(x+h)}-\sqrt{sin2x}}{h}\times \frac{\sqrt{sin2(x+h)}+\sqrt{sin2x}}{\sqrt{sin2(x+h)}+\sqrt{sin2x}}$

$=li{m}_{h\to 0}\frac{sin(2x+2h)-sin2x}{(\sqrt{sin2(x+h)}+\sqrt{sin2x})}$

$=li{m}_{h\to 0}\frac{sin(2x+2h)-sin2x}{(\sqrt{sin(2x+2h)}+\sqrt{sin2x})}[\because sinC-sinD=2cos\frac{C+D}{2}sin\frac{C-D}{2}]$

$=li{m}_{h\to 0}\frac{2cos(2x+h)\times sinh}{h}\times \frac{1}{\sqrt{sin(2x+2h)+\sqrt{sin2x}}}$

$=\frac{2cos2x}{2\sqrt{sin2x}}=\frac{cos2x}{\sqrt{sin2x}}$

(v) We have,

$f\left(x\right)=\frac{sinx}{x}$

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

$=li{m}_{h\to 0}\frac{\frac{sin(x+h)}{x+h}-\frac{sinx}{x}}{h}=li{m}_{h\to 0}\frac{xsin(x+h)-(x+h)sinx}{xh(x+h)}$

$=li{m}_{h\to 0}\frac{x(sinx.cosh+cosx.sinh)-x.sinx-h.sinx}{xh(x+h)}$

$[\because sin(A+B)=sinA.cosB+cosA.sinB]$

$=li{m}_{h\to 0}\frac{x.sinx(cosh-1)}{xh(x+h)}+\frac{x.cosx.sinh}{(x+h)xh}-\frac{hsinx}{(x+h)xh}[\because 1-cosh=2si{n}^2\frac{h}{2}]$

$=\frac{-xsinx}{x(x+h)}\times \frac{2si{n}^2\frac{h}{2}}{\frac{h^2}{2}}\times \frac{h}{4}+\frac{xcosx}{x^2}-\frac{sinx}{x^2}$

$\because h\to 0\Rightarrow \frac{h}{2}\to 0\text{and}li{m}_{\theta \to 0}\frac{sin\theta }{\theta }=1$

$=0+\frac{xcosx-sinx}{x^2}=\frac{xcosx-sinx}{x^2}$

(vi) We have,

$f\left(x\right)=-\frac{cosx}{x}$

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

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

$=li{m}_{h\to 0}\frac{x.cos(x+h)-(x+h)cosx}{(x+h)\times h}[\because cos(A+B)=cosA.cosB-sinA.sinB]$

$=li{m}_{h\to 0}\frac{x[cosx.cosh-sinx.sinh]-x.cosx-h.cosx}{(x+h)x.h}$

$=li{m}_{h\to 0}\frac{xcosx(cosh-1)}{(x+h)x.h}-\frac{x.cosx.sinh}{(x+h)\times h}-\frac{h.cosx}{(x+h)\times h}$

$=li{m}_{h\to 0}\frac{-xcosx.2si{n}^2\frac{h}{2}}{(x+h)\times \frac{h^2}{4}}\times \frac{h^2}{4}-\frac{s.sinx}{x(x+h)}-\frac{cosx}{x(x+h)}$

$=0-\frac{xsinx}{x^2}-\frac{cosx}{x^2}=-\frac{xsinx-cosx}{x^2}$

(vii) We have,

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

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

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

$=li{m}_{h\to 0}\frac{(x^2+h^2+2hx)(sinx.cosh+cosx.sinh)-x^{2}sinx}{h}$

$[\because sin(A+B)=sinA.cosBcosB.sinA]$

$=li{m}_{h\to 0}\frac{x^{2}sinx(cosh-1)}{h}+\frac{h(h+2x)sin.cosh}{h}+(x+h{)}^{2}cosx\frac{sinh}{h}$

$=li{m}_{h\to 0}-x^{2}sinx\times \frac{2si{n}^2\frac{h}{2}}{{\left(\frac{h}{2}\right)}^2}+\frac{h^2}{4}+(h+2x)sinx.cosh+(x+h{)}^{2}cosx$

$=0+(2xsinx+x^{2}cosx)$

$=2xsinx+x^{2}cosx$

(viii) We have,

$f\left(x\right)=\sqrt{sin(3x+1)}$

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

$=li{m}_{h\to 0}\frac{\sqrt{sin3(x+h)+1}-\sqrt{sin(3x+1)}}{h}$

$=li{m}_{h\to 0}\frac{\sqrt{sin(3x+3h)+1}-\sqrt{sin(3x+1)}}{h}\times \frac{\sqrt{sin(3x+3h)+1}+\sqrt{sin(3x+1)}}{\sqrt{sin(3x+3h)+1}+\sqrt{sin(3x+1)}}$

$=li{m}_{h\to 0}\frac{sin(3x+3h+1)-sin(3x+1)}{h(\sqrt{sin(3x+3h)+1}+\sqrt{sin(3x+1)})}$

$=li{m}_{h\to 0}2cos(3x+1+\frac{3h}{2})\times \frac{sin\frac{3h}{2}}{\frac{3h}{2}}\times \frac{3}{2}\times \frac{1}{\sqrt{sin(3x+3h{)}_1}+\sqrt{sin(3x+1)}}$

$=\frac{3cos(3x+1)}{2\sqrt{sin(x+1)}}[li{m}_{h\to 0}\frac{sin\frac{3h}{2}}{\frac{3h}{2}}=1]$

(ix) We have,

$f\left(x\right)=sinx+cosx$

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

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

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

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

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

$\left[\begin{array}{c}\because sinA-sinB=2sin\frac{A-B}{2}cos\frac{A+B}{2}\\ \text{and}cosA-cosB=2sin\frac{A+B}{2}sin\frac{A-B}{2}\end{array}\right]$

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

$=li{m}_{h\to 0}\frac{sinh}{h}\{cos\frac{x+h}{2}-sin(x+\frac{h}{2})\}[\because li{m}_{h\to 0}\frac{sinh}{h}=1]$

$=cosx-sinx$