Question

(i) $\frac{2}{x}$

(ii) $\frac{1}{\sqrt{x}}$

(iii) $\frac{1}{x^3}$

(iv) $\frac{x^2+1}{x}$

(v) $\frac{x^2-1}{x}$

(vi) $\frac{x+1}{x+2}$

(vii) $\frac{x+2}{3x+5}$

(viii) k xⁿ

(ix) $\frac{1}{\sqrt{3-x}}$

(x) x² + x + 3

(xi) (x + 2)³

(xii) x³ + 4x² + 3x + 2

(xiii) (x² + 1) (x − 5)

(xiv) $\sqrt{2x^2+1}$

(xv) $\frac{2x+3}{x-2}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{2}{x+h}}-{\displaystyle \frac{2}{x}}}{h} \\ =\underset{h\to 0}{\lim }\frac{2x-2x-2h}{hx(x+h)} \\ =\underset{h\to 0}{\lim }\frac{-2h}{hx(x+h)} \\ =\underset{h\to 0}{\lim }\frac{-2}{x(x+h)} \\ =\frac{-2}{x^2} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{\sqrt{x+h}}}-{\displaystyle \frac{1}{\sqrt{x}}}}{h} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}\times \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} \\ =\underset{h\to 0}{\lim }\frac{x-x-h}{h\sqrt{x}\sqrt{x+h}\left(\sqrt{x}+\sqrt{x+h}\right)} \\ =\underset{h\to 0}{\lim }\frac{-h}{h\sqrt{x}\sqrt{x+h}\left(\sqrt{x}+\sqrt{x+h}\right)} \\ =\underset{h\to 0}{\lim }\frac{-1}{\sqrt{x}\sqrt{x+h}\left(\sqrt{x}+\sqrt{x+h}\right)} \\ =\frac{-1}{\sqrt{x}\sqrt{x}\left(\sqrt{x}+\sqrt{x}\right)} \\ =\frac{-1}{x\times 2\sqrt{x}} \\ =\frac{-1}{2x^{{\displaystyle \frac{3}{2}}}} \\ =-\frac{1}{2}{x}^{\frac{-3}{2}} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{(x+h{)}^3}}-{\displaystyle \frac{1}{x^3}}}{h} \\ =\underset{h\to 0}{\lim }\frac{x^3-(x+h{)}^3}{h(x+h{)}^3{x}^3} \\ =\underset{h\to 0}{\lim }\frac{x^3-x^3-3x^{2}h-3x{h}^2-h^3}{h(x+h{)}^3{x}^3} \\ =\underset{h\to 0}{\lim }\frac{-3x^{2}h-3x{h}^2-h^3}{h(x+h{)}^3{x}^3} \\ =\underset{h\to 0}{\lim }\frac{h\left(-3x^2-3xh-h^2\right)}{h(x+h{)}^3{x}^3} \\ =\underset{h\to 0}{\lim }\frac{\left(-3x^2-3xh-h^2\right)}{(x+h{)}^3{x}^3} \\ =\frac{-3x^2}{x^6} \\ =\frac{-3}{x^4} \\ =-3x^{-4} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{(x+h{)}^2+1}{x+h}}-{\displaystyle \frac{x^2+1}{x}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{x^2+2xh+h^2+1}{x+h}}-{\displaystyle \frac{x^2+1}{x}}}{h} \\ =\underset{h\to 0}{\lim }\frac{x^3+2x^{2}h+h^{2}x+x-x^3-x^{2}h-x-h}{xh(x+h)} \\ =\underset{h\to 0}{\lim }\frac{x^{2}h+h^{2}x-h}{x(x+h)} \\ =\underset{h\to 0}{\lim }\frac{h(x^2+hx-1)}{xh(x+h)} \\ =\underset{h\to 0}{\lim }\frac{x^2+hx-1}{x(x+h)} \\ =\frac{x^2-1}{x^2} \end{gathered}$$

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

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

$$\begin{gathered} \left(\mathrm{vii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{x+h+2}{3\left(x+h\right)+5}}-{\displaystyle \frac{x+2}{3x+5}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{x+h+2}{3x+3h+5}}-{\displaystyle \frac{x+2}{3x+5}}}{h} \\ =\underset{h\to 0}{\lim }\frac{\left(x+h+2\right)\left(3x+5\right)-\left(3x+3h+5\right)\left(x+2\right)}{h\left(3x+3h+5\right)\left(3x+5\right)} \\ =\underset{h\to 0}{\lim }\frac{3x^2+3xh+6x+5x+5h+10-3x^2-3xh-5x-6x-6h-10}{h\left(3x+3h+5\right)\left(3x+5\right)} \\ =\underset{h\to 0}{\lim }\frac{-h}{h\left(3x+3h+5\right)\left(3x+5\right)} \\ =\underset{h\to 0}{\lim }\frac{-1}{\left(3x+3h+5\right)\left(3x+5\right)} \\ =\frac{-1}{{\left(3x+5\right)}^2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{viii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{k{\left(x+h\right)}^n-k{x}^n}{h} \\ =\underset{\left(x+h\right)-x\to 0}{\lim }\frac{k\left[{\left(x+h\right)}^n-x^n\right]}{\left(x+h\right)-x} \\ \text{Here, we have:} \\ \underset{\mathrm{x}\to \mathrm{a}}{\lim }\frac{x^{\mathrm{m}}-a^{\mathrm{m}}}{x-a}\text{=}\mathit{\text{m}}{\mathit{\text{a}}}^{m-1} \\ =kn{x}^{n-1} \end{gathered}$$

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

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

$$\begin{gathered} \left(\mathrm{xi}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\left(x+h+2\right)}^3-{\left(x+2\right)}^3}{h} \\ =\underset{h\to 0}{\lim }\frac{\left(x+h+2-x-2\right)\left[{\left(x+h+2\right)}^2+\left(x+h+2\right)\left(x+2\right)+{\left(x+2\right)}^2\right]}{h} \\ =\underset{h\to 0}{\lim }\frac{h\left[{\left(x+h+2\right)}^2+\left(x+h+2\right)\left(x+2\right)+{\left(x+2\right)}^2\right]}{h} \\ =\underset{h\to 0}{\lim }\left[{\left(x+h+2\right)}^2+\left(x+h+2\right)\left(x+2\right)+{\left(x+2\right)}^2\right] \\ =\left[{\left(x+0+2\right)}^2+\left(x+0+2\right)\left(x+2\right)+{\left(x+2\right)}^2\right] \\ ={\left(x+2\right)}^2+{\left(x+2\right)}^2+{\left(x+2\right)}^2 \\ =3{\left(x+2\right)}^2 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{xii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^3+4{\left(x+h\right)}^2+3\left(x+h\right)+2-\left(x^3+4x^2+3x+2\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{x^3+3x^{2}h+3x{h}^2+h^3+4x^2+4h^2+8xh+3x+3h+2-x^3-4x^2-3x-2}{h} \\ =\underset{h\to 0}{\lim }\frac{3x^{2}h+3x{h}^2+h^3+4h^2+8xh+3h+2}{h} \\ =\underset{h\to 0}{\lim }\frac{h\left(3x^2+3xh+h^2+4h+8x+3\right)}{h} \\ =\underset{h\to 0}{\lim }\left(3x^2+3xh+h^2+4h+8x+3\right) \\ =3x^2+8x+3 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{xiii}\right)\left(x^2+1\right)\left(x-5\right)=x^3-5x^2+x-5 \\ \frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\left(x+h\right)}^3-5{\left(x+h\right)}^2+\left(x+h\right)-5-\left(x^3-5x^2+x-5\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{x^3+3x^{2}h+3x{h}^2+h^3-5x^2-5h^2-10xh+x+h-5-x^3+5x^2-x+5}{h} \\ =\underset{h\to 0}{\lim }\frac{3x^{2}h+3x{h}^2+h^3-5h^2-10xh+h}{h} \\ =\underset{h\to 0}{\lim }\frac{h\left(3x^2+3xh+h^2-5h-10x+1\right)}{h} \\ =\underset{h\to 0}{\lim }\left(3x^2+3xh+h^2-5h-10x+1\right) \\ =3x^2-10x+1 \end{gathered}$$

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

$$\begin{gathered} \left(\mathrm{xv}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{2\left(x+h\right)+3}{x+h-2}}-{\displaystyle \frac{2x+3}{x-2}}}{h} \\ =\underset{h\to 0}{\lim }\frac{\left(2x+2h+3\right)\left(x-2\right)-\left(x+h-2\right)\left(2x+3\right)}{h\left(x+h-2\right)\left(x-2\right)} \\ =\underset{h\to 0}{\lim }\frac{2x^2+2xh+3x-4x-4h-6-2x^2-2xh+4x-3x-3h+6}{h\left(x+h-2\right)\left(x-2\right)} \\ =\underset{h\to 0}{\lim }\frac{-7h}{h\left(x+h-2\right)\left(x-2\right)} \\ =\underset{h\to 0}{\lim }\frac{-7}{\left(x+h-2\right)\left(x-2\right)} \\ =\frac{-7}{\left(x-2\right)\left(x-2\right)} \\ =\frac{-7}{{\left(x-2\right)}^2} \end{gathered}$$