MATH 180 FALL 1995

ASSIGNMENT #21

4.1, #2. The derivatives of the linear functions

r(t) = 2t − 4 and s(t) = 3r(t) = 6t − 12

are the slopes of their graphs. Thus r′(t) = 2, s′(t) = 6, and s′(t) = 3r′(t). 4.2, #2, 4, 5.

\begin{matrix}y & x^{−12} & x^{3/4} & x^{−3/4} \\ y′ & −12x^{−13} & (3/4)x^{−1/4} & −(3/4)x^{−7/4} \end{matrix}

4.2, #6, 11, 12

\begin{matrix}y & x^{−4} & − 3x^{4} − 4x^{3} − 6x + 2 & 3t^{5} − 5\sqrt{t} + 7/t \\ y′ & −4x^{−5} & −12x^{3} − 12x^{2} − 6 & 15t^{4} − (5/2)t^{−1/2} − 7t^{−2} \end{matrix}

4.2, #30. If f(t) = 2t^{3} − 4t^{2} + 3t − 1, then f′(t) = 6t^{2} − 8t + 3, f′′(t) = 12t − 8.