Question

(i) − x

(ii) (−x)⁻¹

(iii) sin (x + 1)

(iv) $\cos \left(x-\frac{\mathrm{\pi }}{8}\right)$

Answer 1

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

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

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

$$\begin{gathered} \left(\mathrm{iv}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(\cos \left(x-\frac{\mathrm{\pi }}{8}\right)\right)=\underset{h\to 0}{\lim }\frac{\cos \left(x+h-{\displaystyle \frac{\mathrm{\pi }}{8}}\right)-\cos \left(x-{\displaystyle \frac{\mathrm{\pi }}{8}}\right)}{h} \\ \text{We know:} \\ \cos C-\cos D=-2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right) \\ =\underset{h\to 0}{\lim }\frac{-2\sin \left({\displaystyle \frac{x+h-{\displaystyle \frac{\mathrm{\pi }}{8}}+x-{\displaystyle \frac{\mathrm{\pi }}{8}}}{2}}\right)\sin \left({\displaystyle \frac{x+h-{\displaystyle \frac{\mathrm{\pi }}{8}}-x+{\displaystyle \frac{\mathrm{\pi }}{8}}}{2}}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{-2\sin \left({\displaystyle \frac{2x+h-{\displaystyle \frac{\mathrm{\pi }}{4}}}{2}}\right)\sin \left({\displaystyle \frac{h}{2}}\right)}{h} \\ =-2\underset{h\to 0}{\lim }\sin \left(\frac{2x+h-{\displaystyle \frac{\mathrm{\pi }}{4}}}{2}\right)\underset{h\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{h}{2}}\right)}{{\displaystyle \frac{h}{2}}}\times \frac{1}{2} \\ =-2\sin \left(x-\frac{\mathrm{\pi }}{8}\right)\times \frac{1}{2} \\ =-\sin \left(x-\frac{\mathrm{\pi }}{8}\right) \end{gathered}$$