If y = (x – 1)(2x + 3)(sec x); evaluate y'.

Strategy # 1 - multiply out 1st, then differentiate:

y = (2x^2 + x – 3)(sec x) = 2x^2(sec x) + x(sec x) – 3(sec x)

y' = 2x^2(sec x)(tan x) + 4x(sec x) + x(sec x)(tan x) + (sec x) – 3(sec x)(tan x).

Strategy # 2 - use the product rule, then segregate and recombine like terms:

y' = (x – 1)(2x + 3)(sec x)(tan x) + 2(x – 1)(sec x) + (2x + 3)(sec x) =

(2x^2 + x – 3)(sec x)(tan x) + 2x(sec x) – 2(sec x) + 2x(sec x) + 3(sec x) =

2x^2(sec x)(tan x) + x(sec x)(tan x) – 3(sec x)(tan x) + 2x(sec x) – 2(sec x) + 2x(sec x) + 3(sec x) =

2x^2(sec x)(tan x) + 4x(sec x) + x(sec x)(tan x) + (sec x) – 3(sec x)(tan x).

Strategy # 3 - take the natural log of both sides, then differentiate:

ln y = ln(x – 1) + ln(2x + 3) + ln(sec x)

y'/y = [1/(x – 1)] + [2/(2x + 3)] + [(sec x)(tan x)/(sec x)] = [1/(x – 1)] + [2/(2x + 3)] + (tan x)

Next we multiplying both sides by y = (x – 1)(2x + 3)(sec x);

y' = [(x – 1)(2x + 3)(sec x)/(x – 1)] + [2(x – 1)(2x + 3)(sec x)/(2x + 3)] + (x – 1)(2x + 3)(sec x)(tan x) =

(2x + 3)(sec x) + 2(x – 1)(sec x) + (x – 1)(2x + 3)(sec x)(tan x); which is equivalent to the 1st step of strategy # 2.

Which way you choose to evaluate any “dy/dx” will of course depend on the composition of y = f(x)g(x)h(x), etc. And perhaps you will have to experiment in advance of the exam with various combinations of algebraic, trigonometric and logarithmic or exponential functions. Remember, your teacher or professor can be quite “resourceful” come test time!

So, when time is tight on that final, do what works best for YOU!