MATH 180 FALL 1995 10:00 SECTION FONG

ASSIGNMENT #24

4.5, #2.

(\sqrt{1 − x^{2}})′ = \frac{1}{2}(1 − x^{2})^{−1/2}(−2x) = −\frac{x}{\sqrt{1 − x^{2}}}

4.5, #7, 10.

\begin{aligned}y & = e^{3w/2}, \\ y′ & = \frac{3}{2}e^{3w/2} \end{aligned}  \begin{aligned}w & = e^{\sqrt{s}}, \\ w′ & = \frac{1}{2\sqrt{s}}e^{\sqrt{s}} \end{aligned}

4.5, #11, 16.

\begin{aligned}(te^{−t^{2}})′ & = e^{−t^{2}} + te^{−t^{2}}(−2t) = e^{−t^{2}}(1 − 2t^{2}) \\ ((e^{z} + 1)^{−2})′ & = −2(e^{z} + 1)^{3} · e^{z} = −\frac{2e^{z}}{(e^{z} + 1)^{−3}} \end{aligned}

4.5, #19.

((5w^{2} + 3)e^{w^{2}})′ = (5w^{2} + 3)e^{w^{2}} · 2w + (10w)e^{w^{2}} = (10w^{3} + 16w)e^{w^{2}}

4.5, #28. If f(x) = (2x + 1)^{10}(3x − 1)^{7}, then

f′(x) = 10(2x + 1)^{9} · 2(3x − 1)^{7} + (2x + 1)^{10} · 7(3x − 1)^{6}3
= (2x + 1)^{9}(3x − 1)^{6}[20(3x − 1) + 21(2x + 1)]
= (2x + 1)^{9}(3x − 1)^{7}(102x + 1)

We may view f′(x) as the product of three factors, namely, (2x + 1)^{9}, (3x − 1)^{7}, and (102x + 1), and differentiate f′(x) accordingly. Thus

f′′(x) = [(2x + 1)^{9}]′(3x − 1)^{7}(102x + 1) + (2x + 1)^{9}[(3x − 1)^{7}]′(102x + 1)
+ (2x + 1)^{9}(3x − 1)^{7}[102x + 1]′
= (18(2x + 1)^{8})(3x − 1)^{7}(102x + 1) + (2x + 1)^{9}(21(3x − 1)^{6})(102x + 1)
+ (2x + 1)^{9}(3x − 1)^{7}(102)