Question

Let $y=\left|\tan \left(\frac{\mathrm{\pi }}{4}-x\right)\right|,$ Then $\frac{dy}{dx}$ at $x=\frac{\mathrm{\pi }}{4}$ is

Answer 1

Step 1: Find left-hand differentiation:

Given: $y=\left|\tan \left(\frac{\mathrm{\pi }}{4}-x\right)\right|$

Left-hand differentiation

For left hand differentiation $x\to \frac{\mathrm{\pi }}{4}-h$

$f(\frac{\mathrm{\pi }}{4}-h)=\left|\tan \left(\frac{\mathrm{\pi }}{4}-x\right)\right|\to \left|\tan \left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}+h\right)\right|=\tan h$

$$\begin{gathered} =\underset{h\to 0}{\lim }\frac{f(a-h)-f\left(a\right)}{-h} \\ =\underset{h\to 0}{\lim }\frac{f({\displaystyle \frac{\mathrm{\pi }}{4}}-h)-f\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{-h} \\ =\underset{h\to 0}{\lim }\frac{\tan h-\left[\left|\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}}\right)\right|\right]}{-h}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{tan}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{]} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \tan h}-0}{-h} \\ =\underset{h\to 0}{-\lim }\frac{{\displaystyle \tan h}}{h}\mathbf{[}\mathbf{\because }\mathbf{}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{tan}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{]} \\ =-1 \end{gathered}$$

Step 2: Find right-hand differentiation:

Right-hand differentiation

For right hand differentiation $x\to \frac{\mathrm{\pi }}{4}+h$

$f(\frac{\mathrm{\pi }}{4}+h)=\left|\tan \left(\frac{\mathrm{\pi }}{4}-x\right)\right|\to \left|\tan \left(\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}-h\right)\right|=\tan h$

$$\begin{gathered} =\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{f({\displaystyle \frac{\mathrm{\pi }}{4}}+h)-f\left({\displaystyle \frac{\mathrm{\pi }}{4}}\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\tan h-\left[\tan \left({\displaystyle 0}\right)\right]}{h}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{tan}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{]} \\ =\underset{h\to 0}{\lim }\frac{\tan h-0}{h} \\ =\underset{h\to 0}{\lim }\frac{\tan h}{h}\mathbf{[}\mathbf{}\mathbf{\because }\mathbf{}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{tan}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{]} \\ =1 \end{gathered}$$

As RHD$\ne$LHD

Hence, $\frac{dy}{dx}$does not exist at $x=\frac{\mathrm{\pi }}{4}$.