Question

(i) x sin x

(ii) x cos x

(iii) sin (2x − 3)

(iv) $\sqrt{\sin 2x}$

(v) $\frac{\sin x}{x}$

(vi) $\frac{\cos x}{x}$

(vii) x² sin x

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

(ix) sin x + cos x

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\left(x+h\right)\sin \left(x+h\right)-x\sin x}{h} \\ =\underset{h\to 0}{\lim }\frac{\left(x+h\right)\left(\sin x\cos h+\cos x\sin h\right)-x\sin x}{h} \\ =\underset{h\to 0}{\lim }\frac{x\sin x\cos h+x\cos x\sin h+h\sin x\cos h+h\cos x\sin h}{h} \\ =\underset{h\to 0}{\lim }\frac{x\sin x\cos h-x\sin x+x\cos x\sin h+h\sin x\cos h+h\cos x\sin h-x\sin x}{h} \\ =x\sin x\underset{h\to 0}{\lim }\frac{\left(\cos h-1\right)}{h}+x\cos x\underset{h\to 0}{\lim }\frac{\sin h}{h}+\sin x\underset{h\to 0}{\lim }\cos h+\cos x\underset{h\to 0}{\lim }\sin h \\ =x\sin x\underset{h\to 0}{\lim }\frac{-2{\sin }^2{\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h^2}{4}}}\times \frac{h}{4}+x\cos x\left(1\right)+\sin x\left(1\right)+\cos x\left(0\right) \\ =x\sin x\times \frac{-h}{2}+x\cos x\left(1\right)+\sin x\left(1\right)+\cos x\left(0\right) \\ =-2x\sin x\left(\frac{1}{2}\right)\left(0\right)+x\cos x+\sin x \\ =x\cos x+\sin x \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\left(x+h\right)\cos \left(x+h\right)-x\cos x}{h} \\ =\underset{h\to 0}{\lim }\frac{\left(x+h\right)\left(\cos x\cos h-\sin x\sin h\right)-x\cos x}{h} \\ =\underset{h\to 0}{\lim }\frac{x\cos x\cos h-x\sin x\sin h+h\cos x\cos h-h\sin x\sin h-x\cos x}{h} \\ =\underset{h\to 0}{\lim }\frac{x\cos x\cos h-x\cos x-x\sin x\sin h+h\cos x\cos h-h\sin x\sin h}{h} \\ =x\cos x\underset{h\to 0}{\lim }\frac{\left(\cos h-1\right)}{h}-x\sin x\underset{h\to 0}{\lim }\frac{\sin h}{h}+\cos x\underset{h\to 0}{\lim }\cos h+\sin x\underset{h\to 0}{\lim }\sin h \\ =x\cos x\underset{h\to 0}{\lim }\frac{-2{\sin }^2{\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h^2}{4}}}\times \frac{h}{4}-x\sin x\left(1\right)+\cos x\left(1\right)+\sin x\left(0\right) \\ =x\cos x\underset{h\to 0}{\lim }\frac{-h}{2}-x\sin x\left(1\right)+\cos x\left(1\right)+\sin x\left(0\right) \\ =-x\cos x\left(0\right)-x\sin x+\cos x \\ =-x\sin x+\cos x \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\sin \left(2x+2h-3\right)-\sin \left(2x-3\right)}{h} \\ \text{We know:} \\ \text{sin C-sin D=2 cos}\left(\frac{C+D}{2}\right)\text{sin}\left(\frac{C-D}{2}\right) \\ =\underset{h\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{2x+2h-3+2x-3}{2}}\right)\sin \left({\displaystyle \frac{2x+2h-3+2x-3}{2}}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{4x+2h-6}{2}}\right)\sin \left(h\right)}{h} \\ =\underset{h\to 0}{\lim }2\cos \left(\frac{4x+2h-6}{2}\right)\underset{h\to 0}{\lim }\frac{\sin h}{h} \\ =2\cos \left(\frac{4x-6}{2}\right)\left(1\right) \\ =2\cos \left(2x-3\right) \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{\sin \left(2x+2h\right)}-\sqrt{\sin 2x}}{h}\times \frac{\sqrt{\sin \left(2x+2h\right)}+\sqrt{\sin 2x}}{\sqrt{\sin \left(2x+2h\right)}+\sqrt{\sin 2x}} \\ =\underset{h\to 0}{\lim }\frac{\sin \left(2x+2h\right)-\sin 2x}{h\left(\sqrt{\sin \left(2x+2h\right)}+\sqrt{\sin 2x}\right)} \\ \text{We have:} \\ \text{sin C-sin D= 2 cos}\left(\frac{C+D}{2}\right)\text{sin}\left(\frac{C-D}{2}\right) \\ =\underset{h\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{2x+2h+2x}{2}}\right)\sin \left({\displaystyle \frac{2x+2h-2x}{2}}\right)}{h\left(\sqrt{\sin \left(2x+2h\right)}+\sqrt{\sin 2x}\right)} \\ =\underset{h\to 0}{\lim }\frac{2\cos \left(2x+h\right)\sin h}{h\left(\sqrt{\sin \left(2x+2h\right)}+\sqrt{\sin 2x}\right)} \\ =\underset{h\to 0}{\lim }2\cos \left(2x+h\right)\underset{h\to 0}{\lim }\frac{\sin h}{h}\underset{h\to 0}{\lim }\frac{1}{\left(\sqrt{\sin \left(2x+2h\right)}+\sqrt{\sin 2x}\right)} \\ =2\cos 2x\left(1\right)\frac{1}{\sqrt{\sin 2x}+\sqrt{\sin 2x}} \\ =\frac{2\cos 2x}{2\sqrt{\sin 2x}} \\ =\frac{\cos 2x}{\sqrt{\sin 2x}} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{\sin \left(x+h\right)}{x+h}}-{\displaystyle \frac{\sin x}{x}}}{h} \\ =\underset{h\to 0}{\lim }\frac{x\sin \left(x+h\right)-\left(x+h\right)\sin x}{hx\left(x+h\right)} \\ =\underset{h\to 0}{\lim }\frac{x\left(\sin x\cos h+\cos x\sin h\right)-x\sin x-h\sin x}{hx\left(x+h\right)} \\ =\underset{h\to 0}{\lim }\frac{x\sin x\cos h+x\cos x\sin h-x\sin x-h\sin x}{hx\left(x+h\right)} \\ =\underset{h\to 0}{\lim }\frac{x\sin x\cos h-x\sin x+x\cos x\sin h-h\sin x}{hx\left(x+h\right)} \\ =x\sin x\underset{h\to 0}{\lim }\frac{\cos h-1}{h}+\frac{x\cos x}{x}\underset{h\to 0}{\lim }\frac{\sin h}{h}\underset{h\to 0}{\lim }\frac{1}{x+h}-\frac{\sin x}{x}\underset{h\to 0}{\lim }\frac{1}{x+h} \\ =x\sin x\underset{h\to 0}{\lim }\frac{-2{\sin }^2{\displaystyle \frac{h}{2}}}{h}+\frac{x\cos x}{x}\underset{h\to 0}{\lim }\frac{\sin h}{h}\underset{h\to 0}{\lim }\frac{1}{x+h}-\frac{\sin x}{x}\underset{h\to 0}{\lim }\frac{1}{x+h} \\ =x\sin x\underset{h\to 0}{\lim }\frac{-2{\sin }^2{\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h^2}{4}}}\times \frac{h}{4}+\frac{x\cos x}{x}\underset{h\to 0}{\lim }\frac{\sin h}{h}\underset{h\to 0}{\lim }\frac{1}{x+h}-\frac{\sin x}{x}\underset{h\to 0}{\lim }\frac{1}{x+h} \\ =-x\sin x\times \underset{h\to 0}{\lim }\frac{h}{2}+\frac{x\cos x}{x}\underset{h\to 0}{\lim }\frac{\sin h}{h}\underset{h\to 0}{\lim }\frac{1}{x+h}-\frac{\sin x}{x}\underset{h\to 0}{\lim }\frac{1}{x+h} \\ =-x\sin x\left(\frac{1}{2}\right)\left(0\right)+\frac{cosx}{x}-\frac{sinx}{x^2} \\ =\frac{\cos x}{x}-\frac{\sin x}{x^2} \\ =\frac{x\cos x-\sin x}{x^2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{\cos \left(x+h\right)}{x+h}}-{\displaystyle \frac{\cos x}{x}}}{h} \\ =\underset{h\to 0}{\lim }\frac{x\cos \left(x+h\right)-\left(x+h\right)\cos x}{hx\left(x+h\right)} \\ =\underset{h\to 0}{\lim }\frac{x\left(\cos x\cos h-\sin x\sin h\right)-x\cos x-h\cos x}{hx\left(x+h\right)} \\ =\underset{h\to 0}{\lim }\frac{x\cos x\cos h-x\sin x\sin h-x\cos x-h\cos x}{hx\left(x+h\right)} \\ =\underset{h\to 0}{\lim }\frac{x\cos x\cos h-x\cos x-x\sin x\sin h-h\cos x}{hx\left(x+h\right)} \\ =x\cos x\underset{h\to 0}{\lim }\frac{\cos h-1}{h}-\frac{x\sin x}{x}\underset{h\to 0}{\lim }\frac{\sin h}{h}\underset{h\to 0}{\lim }\frac{1}{x+h}-\frac{\cos x}{x}\underset{h\to 0}{\lim }\frac{1}{x+h} \\ =x\cos x\underset{h\to 0}{\lim }\frac{-2{\sin }^2{\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h^2}{4}}}\times \frac{h}{4}-\frac{x\sin x}{x}\underset{h\to 0}{\lim }\frac{\sin h}{h}\underset{h\to 0}{\lim }\frac{1}{x+h}-\frac{\cos x}{x}\underset{h\to 0}{\lim }\frac{1}{x+h}\left[\because \underset{h\to 0}{\lim }\frac{{\sin }^2{\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h^2}{4}}}=\underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\times \underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}=1\times 1,\mathrm{i}.\mathrm{e}.1\right] \\ =-x\cos x\underset{h\to 0}{\lim }\frac{h}{2}-\frac{x\sin x}{x}\underset{h\to 0}{\lim }\frac{\sin h}{h}\underset{h\to 0}{\lim }\frac{1}{x+h}-\frac{\cos x}{x}\underset{h\to 0}{\lim }\frac{1}{x+h} \\ =-x\cos x\times 0-\sin x\left(1\right)\frac{1}{x}-\frac{\cos x}{x}\frac{1}{x} \\ =0-\frac{\sin x}{x}-\frac{\cos x}{x^2} \\ =-\frac{\sin x}{x}-\frac{\cos x}{x^2} \\ =\frac{-x\sin x-\cos x}{x^2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^2\sin \left(x+h\right)-x^2\sin x}{h} \\ =\underset{h\to 0}{\lim }\frac{\left(x^2+h^2+2xh\right)\left(\sin x\cos h+\cos x\sin h\right)-x^2\sin x}{h} \\ =\underset{h\to 0}{\lim }\frac{x^2\sin x\cos h+x^2\cos x\sin h+h^2\sin x\cos h+h^2\cos x\sin h+2xh\sin x\cos h+2xh\cos x\sin h-x^2\sin x}{h} \\ =\underset{h\to 0}{\lim }\frac{x^2\sin x\cos h-x^2\sin x+x^2\cos x\sin h+h^2\sin x\cos h+h^2\cos x\sin h+2xh\sin x\cos h+2xh\cos x\sin h}{h} \\ =x^2\sin x\underset{h\to 0}{\lim }\frac{\cos h-1}{h}+x^2\cos x\underset{h\to 0}{\lim }\frac{\sin h}{h}+\sin x\underset{h\to 0}{\lim }h\cos h+\cos x\underset{h\to 0}{\lim }h\sin h+2x\sin x\underset{h\to 0}{\lim }\cos h+2x\cos x\underset{h\to 0}{\lim }\sin h \\ =x^2\sin x\underset{h\to 0}{\lim }\frac{-2{\sin }^2{\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h^2}{4}}}\times \frac{h}{4}+x^2\cos x\underset{h\to 0}{\lim }\frac{\sin h}{h}+\sin x\underset{h\to 0}{\lim }h\cos h+\cos x\underset{h\to 0}{\lim }h\sin h+2x\sin x\underset{h\to 0}{\lim }\cos h+2x\cos x\underset{h\to 0}{\lim }\sin h\left[\because \underset{h\to 0}{\lim }\frac{{\sin }^2{\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h^2}{4}}}=\underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\times \underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}=1\times 1,\mathrm{i}.\mathrm{e}.1\right] \\ =-x^2\sin x\times \underset{h\to 0}{\lim }\frac{h}{2}+x^2\cos x\underset{h\to 0}{\lim }\frac{\sin h}{h}+\sin x\underset{h\to 0}{\lim }h\cos h+\cos x\underset{h\to 0}{\lim }h\sin h+2x\sin x\underset{h\to 0}{\lim }\cos h+2x\cos x\underset{h\to 0}{\lim }\sin h \\ =-x^2\sin x\times 0+x^2\cos x\left(1\right)+\sin x\left(0\right)+\cos x\left(0\right)+2x\sin x\left(1\right)+2x\cos x\left(0\right) \\ =0+x^2\cos x+2x\sin x \\ =0+x^2\cos x+2x\sin x \\ =x^2\cos x+2x\sin x \end{gathered}$$

$$\begin{gathered} \left(\mathrm{viii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{\sin \left(3\left(x+h\right)+1\right)}-\sqrt{\sin \left(3x+1\right)}}{h} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{\sin \left(3x+3h+1\right)}-\sqrt{\sin \left(3x+1\right)}}{h}\times \frac{\sqrt{\sin \left(3x+3h+1\right)}+\sqrt{\sin \left(3x+1\right)}}{\sqrt{\sin \left(3x+3h+1\right)}+\sqrt{\sin \left(3x+1\right)}} \\ =\underset{h\to 0}{\lim }\frac{\sin \left(3x+3h+1\right)-\sin \left(3x+1\right)}{h\left(\sqrt{\sin \left(3x+3h+1\right)}+\sqrt{\sin \left(3x+1\right)}\right)} \\ \text{We have:} \\ \text{sin C-sin D= 2 cos}\left(\frac{C+D}{2}\right)\text{sin}\left(\frac{C-D}{2}\right) \\ =\underset{h\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{3x+3h+1+3x+1}{2}}\right)\sin \left({\displaystyle \frac{3x+3h+1-3x-1}{2}}\right)}{h\left(\sqrt{\sin \left(3x+3h+1\right)}+\sqrt{\sin \left(3x+1\right)}\right)} \\ =\underset{h\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{6x+3h+2}{2}}\right)\sin {\displaystyle \frac{3h}{2}}}{h\left(\sqrt{\sin \left(3x+3h+1\right)}+\sqrt{\sin \left(3x+1\right)}\right)} \\ =\underset{h\to 0}{\lim }2\cos \left(\frac{6x+3h+2}{2}\right)\underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{3h}{2}}}{h\times {\displaystyle \frac{3}{2}}}\times \frac{3}{2}\times \underset{h\to 0}{\lim }\frac{1}{\left(\sqrt{\sin \left(3x+3h+1\right)}+\sqrt{\sin \left(3x+1\right)}\right)} \\ =2\cos \left(3x+1\right)\times \left(\frac{3}{2}\right)\times \frac{1}{\sqrt{\sin \left(3x+1\right)}+\sqrt{\sin \left(3x+1\right)}} \\ =\frac{3\cos \left(3x+1\right)}{2\sqrt{\sin \left(3x+1\right)}} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ix}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\sin \left(x+h\right)+\mathrm{co}s\left(x+h\right)-\sin x-\cos x}{h} \\ =\underset{h\to 0}{\lim }\frac{\sin \left(x+h\right)-\sin x}{h}+\underset{h\to 0}{\lim }\frac{\cos \left(x+h\right)-\cos x}{h} \\ \text{We have:} \\ \text{sin C-sin D= 2 cos}\left(\frac{C+D}{2}\right)\text{sin}\left(\frac{C-D}{2}\right) \\ \text{And, cos}C-\cos D=-2\text{sin}\left(\frac{C+D}{2}\right)\text{sin}\left(\frac{C-D}{2}\right) \\ =\underset{h\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{2x+h}{2}}\right)\sin {\displaystyle \frac{h}{2}}}{h}+\underset{h\to 0}{\lim }\frac{-2\sin \left({\displaystyle \frac{2x+h}{2}}\right)\sin {\displaystyle \frac{h}{2}}}{h} \\ =2\underset{h\to 0}{\lim }\cos \left(\frac{2x+h}{2}\right)\underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\times \frac{1}{2}-2\underset{h\to 0}{\lim }\sin \left(\frac{2x+h}{2}\right)\underset{h\to 0}{\lim }\frac{\sin {\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h}{2}}}\times \frac{1}{2} \\ =2\cos x\times \frac{1}{2}-2\sin x\times \frac{1}{2} \\ =\cos x-\sin x \end{gathered}$$