Question

Find the equation of the tangent line to the curve at the given point?

y = x cot(x) at the point with x-coordinate= π/4

Answer 1

The equation of the tangent line

`y - (\pi )/(4) = ((-\pi )/(2) +1)( x - (\pi )/(4) )`

Explanation:

Step(i):-

Given function y = x cot (x) ....(i)

Differentiating equation (i) with respective to 'x' , we get

`(dy)/(dx) = x (-Co sec^(2) (x)) +cot(x) (1)`

Step(ii):-

The slope of the tangent line

`(d y)/(d x) = -x Co-sec^(2) (x) +cot x`

`((d y)/(d x) )x_{=(\pi )/(4) } = -(\pi )/(4) Co-sec^(2) ((\pi )/(4) ) +cot (\pi )/(4)`

We will use trigonometry formulas

`Cosec((\pi )/(4) ) = √(2)`

`sec((\pi )/(4) ) = √(2)`

`Cot((\pi )/(4) ) = 1`

Now the slope of the tangent

`(dy)/(dx) =-(\pi )/(4) (√(2))^(2) )+1`

`(dy)/(dx) =-(\pi )/(2) +1`

Step(iii):-

Given

Substitute
`x=(\pi )/(4)` in y = x cot (x)

`y = (\pi )/(4) cot ((\pi )/(4) )`

`y = (\pi )/(4)`

The point of the tangent line
`(x ,y ) = ((\pi )/(4) , (\pi )/(4) )`

The equation of the tangent line

`y - y_(1) = m ( x - x_(1) )`

`y - (\pi )/(4) = ((-\pi )/(2) +1)( x - (\pi )/(4) )`