Question

(i) $\sin \sqrt{2x}$
(ii) $\cos \sqrt{x}$
(iii) $\tan \sqrt{x}$
(iv) tan x²

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Let}f\left(x\right)=\sin \sqrt{2x} \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ f(x+h)=\sin \sqrt{2\left(x+h\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 \sqrt{2x+2h}-\sin \sqrt{2x}}{h} \\ \text{We know:} \\ \text{sin C- sin D=2 sin}\left(\frac{C-D}{2}\right)\text{cos}\left(\frac{C+D}{2}\right) \\ =\underset{h\to 0}{\lim }\frac{2\sin \left(\sqrt{2x+2h}-\sqrt{2x}\right)\cos \left(\sqrt{2x+2h}-\sqrt{2x}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{2\times 2\sin \left({\displaystyle \frac{\sqrt{2x+2h}-\sqrt{2x}}{2}}\right)\cos \left({\displaystyle \frac{\sqrt{2x+2h}+\sqrt{2x}}{2}}\right)}{2h+2x-2x} \\ =\underset{h\to 0}{\lim }\frac{2\times 2\sin \left({\displaystyle \frac{\sqrt{2x+2h}-\sqrt{2x}}{2}}\right)\cos \left({\displaystyle \frac{\sqrt{2x+2h}-\sqrt{2x}}{2}}\right)}{\left(\sqrt{2x+2h}-\sqrt{2x}\right)\sqrt{2x+2h}+\sqrt{2x}} \\ =\underset{h\to 0}{\lim }\frac{2\times 2\sin \left({\displaystyle \frac{\sqrt{2\mathrm{x}+2\mathrm{h}}-\sqrt{2\mathrm{x}}}{2}}\right)\cos \left({\displaystyle \frac{\sqrt{2\mathrm{x}+2\mathrm{h}}-\sqrt{2\mathrm{x}}}{2}}\right)}{2\times \left({\displaystyle \frac{\sqrt{2\mathrm{x}+2\mathrm{h}}-\sqrt{2\mathrm{x}}}{2}}\right)\left(\sqrt{2\mathrm{x}+2\mathrm{h}}+\sqrt{2\mathrm{x}}\right)} \\ =\underset{h\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{\sqrt{2\mathrm{x}+2\mathrm{h}}-\sqrt{2\mathrm{x}}}{2}}\right)}{\left({\displaystyle \frac{\sqrt{2\mathrm{x}+2\mathrm{h}}-\sqrt{2\mathrm{x}}}{2}}\right)}\underset{\mathrm{h}\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{\sqrt{2\mathrm{x}+2\mathrm{h}}-\sqrt{2\mathrm{x}}}{2}}\right)}{\sqrt{2\mathrm{x}+2\mathrm{h}}+\sqrt{2\mathrm{x}}} \\ =1\times \frac{2\cos \sqrt{2\mathrm{x}}}{2\sqrt{2\mathrm{x}}}\left[\because \underset{h\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{\sqrt{2\mathrm{x}+2\mathrm{h}}-\sqrt{2\mathrm{x}}}{2}}\right)}{\left({\displaystyle \frac{\sqrt{2\mathrm{x}+2\mathrm{h}}-\sqrt{2\mathrm{x}}}{2}}\right)}=1\right] \\ =\frac{\cos \sqrt{2\mathrm{x}}}{\sqrt{2\mathrm{x}}} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{ii}\right)Letf\left(x\right)=\cos \sqrt{x} \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ f(x+h)=\cos \sqrt{x+h} \\ \frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\cos \sqrt{x+h}-\cos \sqrt{x}}{h} \\ \mathrm{We}\mathrm{know}: \\ \cos C-\cos D=-2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right) \\ =\underset{h\to 0}{\lim }\frac{-2\sin \left({\displaystyle \frac{\sqrt{x+h}+\sqrt{x}}{2}}\right)\sin \left({\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{2}}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{-2\sin \left({\displaystyle \frac{\sqrt{x+h}+\sqrt{x}}{2}}\right)\sin \left({\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{2}}\right)}{x+h-x} \\ =\underset{h\to 0}{\lim }\frac{-2\sin \left({\displaystyle \frac{\sqrt{x+h}+\sqrt{x}}{2}}\right)\sin \left({\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{2}}\right)}{2\times \left(\sqrt{x+h}+\sqrt{x}\right){\displaystyle \frac{\left(\sqrt{x+h}-\sqrt{x}\right)}{2}}} \\ =\underset{h\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{2}}\right)}{{\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{2}}}\underset{h\to 0}{\lim }\frac{-\sin \left({\displaystyle \frac{\sqrt{x+h}+\sqrt{x}}{2}}\right)}{\sqrt{x+h}+\sqrt{x}} \\ =1\times \frac{-\sin \sqrt{x}}{2\sqrt{x}}\left[\because \underset{h\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{2}}\right)}{{\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{2}}}=1\right] \\ =\frac{-\sin \sqrt{x}}{2\sqrt{x}} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{Let}f\left(x\right)=\tan \sqrt{x} \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ (x+h)=\tan \sqrt{x+h} \\ \frac{d}{dx}\left(f\right(x\left)\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\tan \sqrt{x+h}-\tan \sqrt{x}}{h} \\ =\underset{h\to 0}{\lim }\frac{\sin \left(\sqrt{x+h}-\sqrt{x}\right)}{h\cos \sqrt{x+h}\cos \sqrt{x}}\left[\because \tan A-\tan B=\frac{\sin (A-B)}{\cos A\cos B}\right] \\ =\underset{h\to 0}{\lim }\frac{\sin \left(\sqrt{x+h}-\sqrt{x}\right)}{\left(x+h-x\right)\cos \sqrt{x+h}\cos \sqrt{x}} \\ =\underset{h\to 0}{\lim }\frac{\sin \left(\sqrt{x+h}-\sqrt{x}\right)}{\left(\sqrt{x+h}-\sqrt{x}\right)\left(\sqrt{x+h}-\sqrt{x}\right)\cos \sqrt{x+h}\cos \sqrt{x}} \\ =\underset{h\to 0}{\lim }\frac{\sin \left(\sqrt{x+h}-\sqrt{x}\right)}{\left(\sqrt{x+h}-\sqrt{x}\right)}.\underset{h\to 0}{\lim }\frac{1}{\left(\sqrt{x+h}+\sqrt{x}\right)\cos \sqrt{x+h}\cos \sqrt{x}}\left[\because \underset{h\to 0}{\lim }\frac{\sin \left(\sqrt{x+h}-\sqrt{x}\right)}{\sqrt{x+h}-\sqrt{x}}=1\right] \\ =1\times \frac{1}{2\sqrt{x}\cos \sqrt{x}\cos \sqrt{x}} \\ =\frac{1}{2\sqrt{x}}{\sec }^2\sqrt{x} \end{gathered}$$


$$\begin{gathered} \left(iv\right)\mathrm{Let}f\left(x\right)=\tan x^2 \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ f(x+h)=\tan (x+h{)}^2 \\ \frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\tan (x+h{)}^2-\tan x^2}{h} \\ =\underset{h\to 0}{\lim }\frac{\sin \left({\left(x+h\right)}^2-x^2\right)}{h\cos {\left(x+h\right)}^2\cos x^2}\left[\because \tan A-\tan B=\frac{\sin (A-B)}{\cos A\cos B}\right] \\ =\underset{h\to 0}{\lim }\frac{\sin (x^2+h^2+2hx-x^2)}{h\cos {\left(x+h\right)}^2\cos x^2} \\ =\underset{h\to 0}{\lim }\frac{\sin (h\left(h+2x)\right)}{h\left(h+2x\right)\cos {\left(x+h\right)}^2\cos x^2}\times \left(h+2x\right) \\ =\underset{h\to 0}{\lim }\frac{\sin (h\left(h+2x)\right)}{(h\left(h+2x)\right)}\underset{h\to 0}{\lim }\frac{h+2x}{\cos (x+h{)}^2\cos x^2}\left[\mathrm{As}\underset{h\to 0}{\lim }\frac{\sin (h\left(h+2x)\right)}{(h\left(h+2x)\right)}=1\right] \\ =1\times \frac{2x}{{\cos }^2{x}^2} \\ =2x{\sec }^2{x}^2 \end{gathered}$$