Question

Differentiate each of the following from first principles:

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

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

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

(iv) x² sin x

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

(vi) sin x + cos x

(vii) x² e^x

(viii) $e^{x^2+1}$

(ix) $e^{\sqrt{2x}}$

(x) $e^{\sqrt{ax+b}}$

(xi) $a^{\sqrt{x}}$

(x) $3^{x^2}$

Answer 1

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

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

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

$$\begin{gathered} \left(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{{\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}$$

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

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

$$\begin{gathered} \left(\mathrm{vii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(x^2{e}^x\right)=\underset{h\to 0}{\lim }\frac{(x+h{)}^2{e}^{(x+h)}-x^2{e}^x}{h} \\ =\underset{h\to 0}{\lim }\frac{(x^2+2xh+h^2)e^x{e}^h-x^2{e}^x}{h} \\ =\underset{h\to 0}{\lim }\frac{x^2{e}^x{e}^h+2xh{e}^x{e}^h+h^2{e}^x{e}^h-x^2{e}^x}{h} \\ =\underset{h\to 0}{\lim }\frac{x^2{e}^x{e}^h-x^2{e}^x}{h}+\underset{h\to 0}{\lim }\frac{2xh{e}^x{e}^h}{h}+\underset{h\to 0}{\lim }\frac{h^2{e}^x{e}^h}{h} \\ =\underset{h\to 0}{\lim }\frac{x^2{e}^x\left(e^h-1\right)}{h}+\underset{h\to 0}{\lim }2x{e}^x{e}^h+\underset{h\to 0}{\lim }h{e}^x{e}^h \\ =x^2{e}^x\left(1\right)+2x{e}^x\left(1\right)+0 \\ =x^2{e}^x+2x{e}^x \\ =\left(x^2+2x\right)e^x \end{gathered}$$

$$\begin{gathered} \left(\mathrm{viii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(e^{x^2+1}\right)=\underset{h\to 0}{\lim }\frac{e^{(x+h{)}^2+1}-e^{x^2+1}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{x^2+h^2+2xh+1}-e^{x^2+1}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{x^2+1}{e}^{h^2+2xh}-e^{x^2+1}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{x^2+1}\left(e^{h\left(h+2x\right)}-1\right)}{h}\times \frac{\left(h+2x\right)}{\left(h+2x\right)} \\ =e^{x^2+1}\underset{h\to 0}{\lim }\frac{e^{h\left(h+2x\right)}-1}{h\left(h+2x\right)}\underset{h\to 0}{\lim }\left(h+2x\right) \\ =e^{x^2+1}\left(1\right)\left(2x\right) \\ =2x{e}^{x^2+1} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ix}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(e^{\sqrt{2x}}\right)=\underset{h\to 0}{\lim }\frac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2x}}}{h} \\ =2\underset{h\to 0}{\lim }\frac{e^{\sqrt{2x+2h}}-e^{\sqrt{2x}}}{2x+2h-2x} \\ =2\underset{h\to 0}{\lim }\frac{e^{\sqrt{2x}}\left(e^{\sqrt{2x+2h}-\sqrt{2x}}-1\right)}{{\left(\sqrt{2x+2h}\right)}^2-{\left(\sqrt{2x}\right)}^2} \\ =2e^{\sqrt{2x}}\underset{h\to 0}{\lim }\frac{e^{\sqrt{2x+2h}-\sqrt{2x}}-1}{\left(\sqrt{2x+2h}-\sqrt{2x}\right)\left(\sqrt{2x+2h}+\sqrt{2x}\right)} \\ =2e^{\sqrt{2x}}\underset{h\to 0}{\lim }\frac{e^{\sqrt{2x+2h}-\sqrt{2x}}-1}{\left(\sqrt{2x+2h}-\sqrt{2x}\right)}\underset{h\to 0}{\lim }\frac{1}{\left(\sqrt{2x+2h}+\sqrt{2x}\right)} \\ =2e^{\sqrt{2x}}\left(1\right)\frac{1}{2\sqrt{2x}} \\ =\frac{e^{\sqrt{2x}}}{\sqrt{2x}} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{x}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(e^{\sqrt{ax+b}}\right)=\underset{h\to 0}{\lim }\frac{e^{\sqrt{ax+ah+b}}-e^{\sqrt{ax+b}}}{h} \\ =a\underset{h\to 0}{\lim }\frac{e^{\sqrt{ax+ah+b}}-e^{\sqrt{ax+b}}}{\left(ax+ah+b\right)-\left(ax+b\right)} \\ =a\underset{h\to 0}{\lim }\frac{e^{\sqrt{ax+b}}\left(e^{\sqrt{ax+ah+b}-\sqrt{ax+b}}-1\right)}{{\left(\sqrt{\left(ax+ah+b\right)}\right)}^2-{\left(\sqrt{\left(ax+b\right)}\right)}^2} \\ =a{e}^{\sqrt{ax+b}}\underset{h\to 0}{\lim }\frac{e^{\sqrt{ax+ah+b}-\sqrt{ax+b}}-1}{\left(\sqrt{\left(ax+ah+b\right)}-\sqrt{\left(ax+b\right)}\right)\left(\sqrt{\left(ax+ah+b\right)}+\sqrt{\left(ax+b\right)}\right)} \\ =a{e}^{\sqrt{ax+b}}\underset{h\to 0}{\lim }\frac{e^{\sqrt{ax+ah+b}-\sqrt{ax+b}}-1}{\sqrt{\left(ax+ah+b\right)}-\sqrt{\left(ax+b\right)}}\underset{h\to 0}{\lim }\frac{1}{\sqrt{\left(ax+ah+b\right)}+\sqrt{\left(ax+b\right)}} \\ =a{e}^{{}^{\sqrt{ax+b}}}\left(1\right)\frac{1}{2\sqrt{ax+b}} \\ =\frac{a{e}^{{}^{\sqrt{ax+b}}}}{2\sqrt{ax+b}} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{xi}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(a^{\sqrt{x}}\right)=\underset{h\to 0}{\lim }\frac{a^{\sqrt{x+h}}-a^{\sqrt{x}}}{h} \\ =\underset{h\to 0}{\lim }\frac{a^{\sqrt{x}}\left(a^{\sqrt{x+h}-\sqrt{x}}-1\right)}{\left(x+h\right)-\left(x\right)} \\ =a^{\sqrt{x}}\underset{h\to 0}{\lim }\frac{\left(a^{\sqrt{x+h}-\sqrt{x}}-1\right)}{{\left(\sqrt{x+h}\right)}^2-{\left(\sqrt{x}\right)}^2} \\ =a^{\sqrt{x}}\underset{h\to 0}{\lim }\frac{\left(a^{\sqrt{x+h}-\sqrt{x}}-1\right)}{\left(\sqrt{x+h}-\sqrt{x}\right)\left(\sqrt{x+h}+\sqrt{x}\right)} \\ =a^{\sqrt{x}}\underset{h\to 0}{\lim }\frac{\left(a^{\sqrt{x+h}-\sqrt{x}}-1\right)}{\left(\sqrt{x+h}-\sqrt{x}\right)}\underset{h\to 0}{\lim }\frac{1}{\left(\sqrt{x+h}+\sqrt{x}\right)} \\ =a^{\sqrt{x}}{\log }_{e}a\frac{1}{2\sqrt{x}} \\ =\frac{1}{2\sqrt{x}}{a}^{\sqrt{x}}{\log }_{e}a \end{gathered}$$

$$\begin{gathered} \left(\mathrm{xii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(3^{x^2}\right)=\underset{h\to 0}{\lim }\frac{3^{{\left(x+h\right)}^2}-3^{x^2}}{h} \\ =\underset{h\to 0}{\lim }\frac{3^{x^2+2xh+h^2}-3^{x^2}}{h} \\ =\underset{h\to 0}{\lim }\frac{3^{x^2}\left(3^{x^2+2xh+h^2-x^2}-1\right)}{h}\times \frac{\left(h+2x\right)}{\left(h+2x\right)} \\ =3^{x^2}\underset{h\to 0}{\lim }\frac{3^{h\left(h+2x\right)}-1}{h\left(h+2x\right)}\underset{h\to 0}{\lim }\left(h+2x\right) \\ =3^{x^2}\log 3\left(2x\right) \\ =2x{3}^{x^2}\log 3 \end{gathered}$$