# Find the equation of the tangent line to the graph of the function f(x) = \sqrt{\arccos(x-1)} ...

## Question:

Find the equation of the tangent line to the graph of the function {eq}f(x) = \sqrt{\arccos(x-1)} {/eq} at the point {eq}(1, 2\pi) {/eq}

## Tangent Line

A line that touches the graph of a function at a single point is called a tangent line. The equation of the tangent line passing through the given point(a,b) and having slope m is given by

{eq}(y-b)=m(x-a) {/eq}

We have {eq}f(x) = \sqrt{\arccos(x-1)} {/eq}

To find the slope of the tangent line, differentiating the given function with respect to the variable x, we get,

{eq}\frac{df(x)}{dx}=\frac{1}{2\sqrt{\arccos(x-1)}}\times \frac{-1}{\sqrt{1-(x-1)^2}}\times 1 \\ \frac{df(x)}{dx}=\frac{1}{2\sqrt{\arccos(x-1)}}\times \frac{-1}{\sqrt{1-(x-1)^2}} {/eq}

at x=1, the slope of the tangent line is

{eq}\frac{df(x)}{dx}=\frac{1}{2\sqrt \arccos(0)}\times -1 \\ \frac{df(x)}{dx}=\frac{-1}{2\sqrt{\frac{\pi}{2}}} \\ \frac{df(x)}{dx}=\frac{-1}{\sqrt{2\pi} } {/eq}

Now the equation of the tangent line with slope {eq}\frac{-1}{\sqrt{2\pi}} {/eq} and passing through the point {eq}(1,2 \pi) {/eq} is given by

{eq}(y- 2 \pi)=\frac{-1}{\sqrt 2 \pi}(x-1) {/eq}

On solving the above equation, we get,

{eq}y\sqrt 2 \pi - (2 \pi)^{\frac{3}{2}}=-x+1 {/eq}

This implies {eq}x+ y\sqrt 2 \pi=1+(2 \pi)^{\frac{3}{2}} {/eq}

Therefore, the equation of the tangent line is {eq}x+ y\sqrt 2 \pi=1+(2 \pi)^{\frac{3}{2}} {/eq} 