Answer: (given assumed typo corrections)
(V ∘ X)'(t) = 0.06(0.01t+3.6)^2 cm^3/sec.
The rate of change of the volume of the cuboid in change of volume per change in seconds, after t seconds. Not a constant, for good reason.
Part B) y'(x+Δx/2)×Δx gives exactly the same as y(x+Δx)-y(x), 0.3808, since y is quadratic in x so y' is linear in x.
Explanation:
This problem has typos. Assuming:
Cuboid has square [base with side] X cm and height 2X cm [not cm^2]. Total surface area of cuboid is 129.6 cm^2, and X [is] increas[ing] at rate 0.01 cm/sec.
129.6 cm^2 = 2(base cm^2) + 4(side cm^2)
= 2(X cm)^2 + 4(X cm)(2X cm)
= (2X^2 + 8X^2)cm^2
= 10X^2 cm^2
X^2 cm^2 = 129.6/10 = 12.96 cm^2
X cm = √12.96 cm = 3.6 cm
so X(t) = (0.01cm/sec)(t sec) + 3.6 cm, or, omitting units,
X(t) = 0.01t + 3.6
= the length parameter after t seconds, in cm.
V(X) = 2X^3 cm^3
= the volume when the length parameter is X.
dV(X(t))/dt = (dV(X)/dX)(X(t)) × dX(t)/dt
that is, (V ∘ X)'(t) = V'(X(t)) × X'(t) chain rule
V'(X) = 6X^2 cm^3/cm
= the rate of change of volume per change in length parameter when the length parameter is X, units cm^3/cm. Not a constant (why?).
X'(t) = 0.01 cm/sec
= the rate of change of length parameter per change in time parameter, after t seconds, units cm/sec.
V(X(t)) = (V ∘ X)(t) = 2(0.01t+3.6)^3 cm^3
= the volume after t seconds, in cm^3
V'(X(t)) = 6(0.01t+3.6)^2 cm^2
= the rate of change of volume per change in length parameter, after t seconds, in units cm^3/cm.
(V ∘ X)'(t) = ( 6(0.01t+3.6)^2 cm^3/cm )(0.01 cm/sec) = 0.06(0.01t+3.6)^2 cm^3/sec
= the rate of change of the volume per change in time, in cm^3/sec, after t seconds.
Problem to ponder: why is (V ∘ X)'(t) not a constant? Does the change in volume of a cube per change in side length depend on the side length?
Question part b)
Given y=2x²+3x, use differentiation to find small change in y when x increased from 4 to 4.02.
This is a little ambiguous, but "use differentiation" suggests that we want y'(4.02) yunit per xunit, rather than Δy/Δx = (y(4.02)-y(4))/(0.02).
Neither of those make much sense, so I think we are to estimate Δy given x and Δx, without evaluating y(x) at all.
Then we want y'(x+Δx/2)×Δx
y(x) = 2x^2 + 3x
y'(x) = 4x + 3
y(4) = 44
y(4.02) = 44.3808
Δy = 0.3808
Δy/Δx = (0.3808)/(0.02) = 19.04
y'(4) = 19
y'(4.01) = 19.04
y'(4.02) = 19.08
Estimate Δy = (y(x+Δx)-y(x)/Δx without evaluating y() at all, using only y'(x), given x = 4, Δx = 0.02.
y'(x+Δx/2)×Δx = y'(4.01)×0.02 = 19.04×0.02 = 0.3808.
In this case, where y is quadratic in x, this method gives Δy exactly.