Sensitivity at stimulus level x?

Question

Suppose that the relationship between R, the area of the pupil (in square millimeters), and x, the brightness of the light source (in lumen), is given by R = (40 + 23.7x^4)/(1 + 3.95x^4). The rate of change dR/dx is called the sensitivity at stimulus level x.

a) show that R decreases from 40 to 6 as x increases without bound
b) find a formula for the sensitivity as a function of x.
c) Using the result of part (b), plot the graph of the sensitivity as a function of x for x ≥ 0.

Answer 1

R = (40 + 23.7x^4)/(1 + 3.95x^4)

a) When x = 0, R = 40 / 1 = 40

Divide each term on the RHS by x^4 :

R = (40 / x^4 + 23.7) / (1 / x^4 + 3.95)

As x tends to infinity, 40 / x^4 and 1 / x^4 tend to zero.
Therefore, R tends to 23.7 / 3.95 = 6

b) Let u = 40 + 23.7x^4 and v = 1 + 3.95x^4

Then, dR /dx = [v(du/dx) - u(dv/dx)] / v^2

= [(1 + 3.95x^4)(94.8x^3) - (40 + 23.7x^4)(15.8x^3)] / (1 + 3.95x^4)^2

= (94.8x^3 + 374.46x^7 - 632x^3 - 374.46x^7) / (1 + 3.95x^4)^2

= - 537.2x^3 / (1 + 3.95x^4)^2

Therefore, Sensitivity = - 537.2x^3 / (1 + 3.95x^4)^2

c) I've already plotted the graph, so I can say that relevant
values of x are from - 5 to + 5, with a few more values between
-1 and + 1, such as ± 0.2, ± 0.4, ± 0.6 and ± 0.8.

The 2nd derivative is x^2(9750.18x^4 - 1611.6) / (1 + 3.95x^4)^3

When this is zero, x = ± 0.6376188627

Plugging these back into the sensitivity equation gives :
(i) a maximum at (- 0.637618827, + 50.97182957)
(ii) a minimum at (+ 0.637618827, - 50.97182957)

There is also a zero and point of inflexion at the origin.

The two ends of the graph tend to
± infinity and are asymptotic to y = 0.

Answer 2

What kind of math is this, is it relationships and functions from University?