MATH 180 FALL 1995 10:00 SECTION FONG

ASSIGNMENT #23

4.4, #1. Expanding f(x) = x^{2}(x^{3} + 5) to x^{5} + 5x^{2} and then differentiating gives f′(x) = 5x^{4} + 10x. Differentiating f(x) by the product rule gives f′(x) = 2x(x^{3} + 5) + x^{2}(3x^{2}), which simplifies to 5x^{4} + 10x. The two methods give the same answer.

4.4, #3, 4.

\begin{aligned}(xe^{x})′ & = e^{x} + xe^{x} = e^{x}(1 + x), \\ (xe^{−x})′ & = e^{−x} + x(e^{−1})(−1) = e^{−x}(1 − x). \end{aligned}

4.4, #10.

[(t^{2} + 3)e^{t}]′ = (t^{2} + 3)e^{t} + (2t)e^{t} = (t^{2} + 2t + 3)e^{t}.

4.4, #13.

(\frac{t^{2} + 5t + 2}{t + 3})′ = \frac{(t + 3)(2t + 5) − (t^{2} + 5t + 2)}{(t + 3)^{2}} = \frac{t^{2} + 6t + 13}{(t + 3)^{2}}

4.4, #16.

(\frac{\sqrt{t}}{t^{2} + 1})′ = \frac{(t^{2} + 1)\frac{1}{2}t^{− \frac{1}{2}} − (\sqrt{t})2t}{(t^{2} + 1)^{2}}

= \frac{(t^{2} + 1) − 4t^{2}}{2(t^{2} + 1)^{2}\sqrt{t}}

= \frac{1 − 3t^{2}}{2(t^{2} + 1)^{2}\sqrt{t}}

4.5, #4, #8.

\begin{aligned}w & = (t^{3} + 1)^{100}, \\ y & = e^{−4t}, \end{aligned}  \begin{aligned}w′ & = 100(t^{3} + 1)^{99} · 3t^{2} = 300(t^{3} + 1)^{99} · t^{2} \\ y′ & = −4e^{−4t} \end{aligned}