= f′(x) dx. For this problem, f (x) = V = πr^3 , f′(x) = = 4πr^2 = 100π, and dx = dr = ±0.1.
When these values are input, the equation is dy = dV = 100π(±0.1) = ±10π cm^3.
- D In order to approximate the change in the surface area of the cube, use the approximation
formula: dy = f′(x) dx. For this problem, f(x) = A = 6s^2 , f′(x) = = 12s = 120, and dx = ds =
−0.2. When these values are input, the equation is dy = dA = 120(−0.2) = −24cm^2.
- C We need to use implicit differentiation to find .
3 x^2 y + x^3 + y^3 + 3xy^2 = 0
Now, in order to isolate , we move all of the terms that do not contain to the right side of
the equals sign.
x^3 + 3xy^2 = −3x^2 y − y^3
Factor out .
(x^3 + 3xy^2 ) = −3x^2 y − y^3
Divide both sides by (x^3 + 3xy^2 ) to isolate .
= −
- B We can use u-substitution to evaluate the integral.
Let u = 3x^2 and du = 6x dx. If we solve the second term for x dx, we get
du = x dx
Now we can rewrite the integral as
