Okay, let's start.
First, let's differentiate the function f(x) = 1/√(4x² + 1) to find the slope of the tangent line at x = 2. A tangent line represents the instantaneous rate of change of the function at a certain point, which is also the derivative of the function at that point.
The derivative of f(x), which we will denote as f'(x), gives us the slope of the tangent line to the curve at any point x. If we substitute x = 2 into f'(x), we can get the slope of the tangent line at x = 2.
After conducting the differentiation, the slope of the tangent line at x = 2 is obtained.
Next, to find the slope of the normal line, which is a line perpendicular to the tangent line at the given point, we need to use the fact that the slope of a normal line is the negative reciprocal of the slope of the tangent line.
So, we compute our slope of the normal line with the equation `-1/slope_tangent`.
Now, to find the equation of the normal line, we need the y-coordinate of the point of tangency. Substitute x = 2 into the original function f(x) to find the y-coordinate.
So, the y-coordinate at x = 2 is f(2).
Putting all these together, the equation of the normal line can be expressed as y = `slope_normal`*(x - 2) + `f(2)`.
Substituting the previous results, the equation of the normal line is y = 17*sqrt(17)/8*(x - 2) + sqrt(17)/17.
That's it! We've found the equation of the normal line to the curve at x = 2 using calculus! Welcome to the world of differentiation.