16. D Use u-substitution where u = 3x^3 and du = 9x^2 . Therefore, ∫ 18 x^2 sec^2 (3x^3 ) dx = 2 ∫ sec^2 u du
= 2 tan u + C = 2 tan (3x^3 ) + C.
- A. First, plug x = 1 into the equation and solve for y; y = 5. Next, take the first derivative of y:
= 3x^2 + 4x −5. Solve for at x = 1: = 2. The slope of the normal line is − . Thus, the
equation of the normal line is y − 5 = − (x − 1). The equation simplified is y = − x + .
- B If we take the limit as x goes to , we get an indeterminate form , so let’s use L’Hôpital’s
Rule. We take the derivative of the numerator and the denominator and we get =
= . Now we can use the double angle formula to simplify the limit. We
get = = . Now, when we take the limit we get
= .
- B Using implicit differentiation, you can evaluate this equation: −2 cos x + 2 sin y cos y = .
After simplifying, = .
- D Via the Chain Rule, f′(x) = 3e^3 x and f′′(x) = 9e^3 x. Plugging in ln 3 for x, results in f′′(ln 3) =
9 e3(ln 3) = 9eln 3
3
= 9eln 27 = 9 · 27 = 243.
- C Use implicit differentiation and plug in for the point (1, 1):
4 y − 6 = 4x^3 + 6x^2 − 2
