Question

16) Find the derivative of cos x² by first principal .
OR
If $f\left(x\right)=\frac{secx-1}{secx+1},provethat\frac{f\left(x\right)}{f\left(x\right)}=\frac{\sin x}{2}$

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{y}=\cos {\mathrm{x}}^2 \\ \Rightarrow \mathrm{y}+∆\mathrm{y}=\cos {\left(\mathrm{x}+∆\mathrm{x}\right)}^2 \\ \Rightarrow ∆\mathrm{y}=\cos {\left(\mathrm{x}+∆\mathrm{x}\right)}^2-\mathrm{y} \\ \Rightarrow ∆\mathrm{y}=\cos {\left(\mathrm{x}+∆\mathrm{x}\right)}^2-\cos {\mathrm{x}}^2 \\ \Rightarrow ∆\mathrm{y}=-2\sin \left[\frac{{\left(\mathrm{x}+∆\mathrm{x}\right)}^2+{\mathrm{x}}^2}{2}\right].\sin \left[\frac{{\left(\mathrm{x}+∆\mathrm{x}\right)}^2-{\mathrm{x}}^2}{2}\right] \\ \Rightarrow \frac{∆\mathrm{y}}{∆\mathrm{x}}=-2\sin \left[\frac{{\left(\mathrm{x}+∆\mathrm{x}\right)}^2+{\mathrm{x}}^2}{2}\right].\frac{\sin \left[{\displaystyle \frac{{\left(∆\mathrm{x}\right)}^2+2\mathrm{x}∆\mathrm{x}}{2}}\right]}{∆\mathrm{x}} \\ \Rightarrow \frac{∆\mathrm{y}}{∆\mathrm{x}}=-2\sin \left[\frac{{\left(\mathrm{x}+∆\mathrm{x}\right)}^2+{\mathrm{x}}^2}{2}\right].\frac{\sin \left[{\displaystyle \frac{∆\mathrm{x}\left(∆\mathrm{x}+2\mathrm{x}\right)}{2}}\right]}{{\displaystyle \frac{∆\mathrm{x}\left(∆\mathrm{x}+2\mathrm{x}\right)}{2}}}\times \frac{\left(∆\mathrm{x}+2\mathrm{x}\right)}{2} \\ \Rightarrow \underset{∆\mathrm{x}\to 0}{\lim }\frac{∆\mathrm{y}}{∆\mathrm{x}}=-2\underset{∆\mathrm{x}\to 0}{\lim }\sin \left[\frac{{\left(\mathrm{x}+∆\mathrm{x}\right)}^2+{\mathrm{x}}^2}{2}\right].\underset{∆\mathrm{x}\to 0}{\lim }\frac{\sin \left[{\displaystyle \frac{∆\mathrm{x}\left(∆\mathrm{x}+2\mathrm{x}\right)}{2}}\right]}{{\displaystyle \frac{∆\mathrm{x}\left(∆\mathrm{x}+2\mathrm{x}\right)}{2}}}.\underset{∆\mathrm{x}\to 0}{\lim }\frac{\left(∆\mathrm{x}+2\mathrm{x}\right)}{2} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-2\sin {\mathrm{x}}^2\times 1\times \mathrm{x} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-2\mathrm{x}\sin {\mathrm{x}}^2 \end{gathered}$$