Let A = πr^2 . Then our formula says that dA = A′dr, where A′ is the derivative of the area with respect to r,
and dr = 0.04 (the change). First, find the derivative of the area: A′ = 2πr. Now, plug in to the formula.
dA = 2πr dr = 2π (3)(0.04) = 0.754The actual change in the area is from 9π to 9.2416π, which is approximately 0.759. As you can see, this
approximation formula is pretty accurate.
Here are some sample problems involving this differential formula. Try them out, then check your work
against the answers directly beneath.
PROBLEM 1. Use differentials to approximate (3.98)^4.
Answer: Let f(x) = x^4 , x = 4, and ∆x = −0.02. Next, find f′(x), which is: f′(x) = 4x^3.
Now, plug in to the formula.
f(x + ∆x) ≈ f(x) + f′(x)∆x(x + ∆x)^4 ≈ x^4 + 4x^3 ∆xIf you plug in x = 4 and ∆x = −0.02, you get
(3.98)^4 ≈ 4^4 + 4(4)^3 (−0.02) ≈ 250.88Check (3.98)^4 by using your calculator; you should get 250.9182722. Not a bad approximation.
You’re probably asking yourself, why can’t I just use my calculator every time? Because most math
teachers are dedicated to teaching you several complicated ways to calculate things without your
calculator.
PROBLEM 2. Use differentials to approximate sin 46°.
Answer: This is a tricky question. The formula doesn’t work if you use degrees. Here’s why: Let f(x) =
sin x, x = 45°, and ∆x = 1°. The derivative is f′(x) = cos x.
If you plug this information into the formula, you get: sin 46° ≈ sin 45° + cos 45°(1°) = . You should
recognize that this is nonsense for two reasons: (1) the sine of any angle is between −1 and 1; and (2) the
answer should be close to sin 45° = .
What went wrong? You have to use radians! As we mentioned before, angles in calculus problems are
measured in radians, not degrees.
Let f(x) = sin x, x = , and ∆x = . Now plug in to the formula.