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 will be `y = (2 - pi)/2x+ pi^2/8`

Explanation:

We know that

`y(pi/4) = pi/4cot(pi/4) = pi/4(1) = pi/4`

Now taking the derivative using the product rule, we have:

`y' = cotx - xcsc^2x`

We now must find the slope of the tangent.

`y'(pi/4) = cot(pi/4) - pi/4(1/sin(pi/4))^2`

`y'(pi/4) = 1 - pi/4(1/(1/sqrt(2)))^2`

`y'(pi/4) = 1 - pi/2`

Now using point-slope form:

`y -y_1 = m(x - x_1)`

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

`y = x - pi/2x - pi/4 + pi^2/8 + pi/4`

`y = (2 - pi)/2x + pi^2/8`

Here is graphical representation of the problem.

As you can see the graph confirms our answer.

Hopefully this helps!