Question

Find the derivatives of the following functions at the indicated points:

(i) sin x at x = $\frac{\mathrm{\pi }}{2}$

(ii) x at x = 1

(iii) 2 cos x at x = $\frac{\mathrm{\pi }}{2}$

(iv) sin 2x at x = $\frac{\mathrm{\pi }}{2}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}: \\ f\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{h\to 0}{\lim }\frac{f\left(\frac{\mathrm{\pi }}{2}+h\right)-f\left(\frac{\mathrm{\pi }}{2}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{sin\left(\frac{\mathrm{\pi }}{2}+h\right)-sin\left(\frac{\mathrm{\pi }}{2}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{cosh-1}{h} \\ \underset{h\to 0}{=\lim }\frac{-2{\sin }^2{\displaystyle \frac{h}{2}}}{h} \\ \underset{}{=\underset{h\to 0}{\lim }\frac{-2{\sin }^2{\displaystyle \frac{h}{2}}}{{\displaystyle \frac{h^2}{4}}}}\times \frac{h}{4} \\ \underset{}{=\underset{h\to 0}{\lim }-1}\times \frac{h}{2} \\ =0 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right)\hspace{0.17em}\mathrm{We}\mathrm{have}: \\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(1+h)-f\left(1\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{1+h-1}{h} \\ =\underset{h\to 0}{\lim }1 \\ =1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}: \\ f\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{h\to 0}{\lim }\frac{f\left(\frac{\mathrm{\pi }}{2}+h\right)-f\left(\frac{\mathrm{\pi }}{2}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{2cos\left(\frac{\mathrm{\pi }}{2}+h\right)-cos\left(\frac{\mathrm{\pi }}{2}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{-2sinh-0}{h} \\ =-2\underset{h\to 0}{\lim }\frac{\sin h}{h} \\ =-2\left(1\right) \\ =-2 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{We}\mathrm{have}: \\ f\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\underset{h\to 0}{\lim }\frac{f\left(\frac{\mathrm{\pi }}{2}+h\right)-f\left(\frac{\mathrm{\pi }}{2}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\sin 2\left(\frac{\mathrm{\pi }}{2}+h\right)-\sin 2\left(\frac{\mathrm{\pi }}{2}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\sin (\mathrm{\pi }+2h)-0}{h} \\ =\underset{h\to 0}{\lim }\frac{-\sin 2h}{h}\times \frac{2}{2} \\ =\underset{h\to 0}{\lim }-\frac{\sin 2h}{2h}\times 2 \\ =-2 \end{gathered}$$