If f(x) = 9 cos2(x), compute its differential df. df = (−18cos(x)sin(x))dx correct: your answer is correct. approximate the change in f when x changes from x = π 6 to x = π 6 + 0.1. (round your answer to three decimal places.) δf = .738 incorrect: your answer is incorrect. approximate the relative change in f as x undergoes this change. (round your answer to three decimal places.)

Given: f(x) = 9 cos (2x)

The differential equation is df = - 18 sin(2x) dx

When x contrasts from π/6 to π/6 + 01, then dx = 0.1.
The variation in f is δf = - 18 sin(π/3) *(0.1) = -1.5588 ≈ -1.559

The computation in the change in f directly:

f(π/6) = 9 cos(π/3) = 4.5
f(π/6 + 0.1) = 9 cos(π/3 + 0.2) = 2.6818
δf = 2.6818- 4.5 = -1.6382 ≈ -1.638

Direct computation of δf is near to the real value but in error. 
The two outcomes will be closer as dx gets smaller.

Answer would be:
δf = -1.559  (correct answer)
δf = -1.638 (approximate answer)