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MATH 180\hfill FALL 1995

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10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#24}

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4.5, \#2.
$$\bigl(\sqrt{1-x^2}\bigr)'=
  {1\over2}(1-x^2)^{-1/2}(-2x)=-{x\over\sqrt{1-x^2}}$$

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4.5, \#7, 10.
$$\eqalign{
    y&=e^{3w/2}, \cr
    y'&={3\over2}e^{3w/2}\cr}\qquad
  \eqalign{
    w&=e^{\sqrt{s}}, \cr
    w'&={1\over2\sqrt s}e^{\sqrt{s}}\cr}$$

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4.5, \#11, 16.
$$\eqalign{
   \bigl(te^{-t^2}\bigr)'&=e^{-t^2}+te^{-t^2}(-2t)
       =e^{-t^2}(1-2t^2)\cr
  \noalign{\vskip4pt}
  \bigl((e^z+1)^{-2}\bigr)'
    &=-2(e^z+1)^3\cdot e^z=-{2e^z\over (e^z+1)^{-3}}\cr}$$

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4.5, \#19.
$$\left((5w^2+3)e^{w^2}\right)'
   =(5w^2+3)e^{w^2}\cdot2w+(10w)e^{w^2}
   =(10w^3+16w)e^{w^2}$$

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4.5, \#28.  If $f(x)=(2x+1)^{10}(3x-1)^7$, then
$$\eqalign{
    f'(x)
      &=10(2x+1)^9\cdot 2(3x-1)^7+ (2x+1)^{10}\cdot 7(3x-1)^6 3\cr
      &=(2x+1)^9 (3x-1)^6[20(3x-1)+21(2x+1)]\cr
      &=(2x+1)^9 (3x-1)^7(102x+1)\cr}$$
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
$$\eqalign{
    f''(x)
       &=[(2x+1)^9]'(3x-1)^7(102x+1)+
          (2x+1)^9 [(3x-1)^7]'(102x+1)\cr
       &\qquad+(2x+1)^9 (3x-1)^7[102x+1]'\cr
       &=(18(2x+1)^8)(3x-1)^7(102x+1)+
          (2x+1)^9 (21(3x-1)^6)(102x+1)\cr
       &\qquad+(2x+1)^9 (3x-1)^7(102)\cr}$$


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