Question

The slope of the tangent to locus of $y=co{s}^{-1}\left(cosx\right)$ at $x=\frac{\mathrm{\pi }}{4}$

Answer 1

Compute the slope of the tangent to the locus

Concept:
If we let $\mathrm{\Delta x}$and$\mathrm{\Delta y}$ be the distances (along the$\mathrm{x}\mathrm{and}\mathrm{y}$ axes, respectively) between two points on a curve, then the slope given by the above definition, $\mathrm{m}=\frac{\mathrm{\Delta y}}{\mathrm{\Delta x}}$

Formula Used:

$\begin{array}{rl}& {\cos }^{-1}(\cos x)=-x,-\pi \le x\le 0\\ & {\cos }^{-1}(\cos x)=x,0<x\le \pi \end{array}$
Calculation:

$y={\cos }^{-1}(\cos x)$
Since,$x=\frac{\mathrm{\pi }}{4}$ which is belongs to $0<x\le \pi$ 0

$\begin{array}{cc}& \text{So,}y=x\\ & \Rightarrow \frac{dy}{dx}=1\end{array}$

Hence,the slope of the tangent of $y={\cos }^{-1}\left(\cos x\right)$ at $x=\frac{\pi }{4}$ is $\mathbf{1}$.