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MATH 180\hfill FALL 1995

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10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#23}

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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.

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4.4, \#3, 4.
$$\eqalign{
  (xe^x)'&=e^x+xe^x=e^x(1+x),\cr
  (xe^{-x})'&=e^{-x}+x(e^{-1})(-1)  =e^{-x}(1-x).\cr}$$

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4.4, \#10.
$$[(t^2+3)e^t]'=(t^2+3)e^t+(2t)e^t=(t^2+2t+3)e^t.$$

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4.4, \#13.
$$\left({t^2+5t+2\over t+3}\right)'=
  {(t+3)(2t+5)-(t^2+5t+2)\over (t+3)^2}=
  {t^2+6t+13\over (t+3)^2}$$

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4.4, \#16.
$$\eqalign{
   \left({\sqrt t\over t^2+1}\right)'
       &={(t^2+1){1\over2}t^{-{1\over2}}-(\sqrt t)2t\over (t^2+1)^2}\cr
       &={(t^2+1)-4t^2\over 2(t^2+1)^2\sqrt t}\cr
       &={1-3t^2\over 2(t^2+1)^2\sqrt t}\cr}$$

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4.5, \#4, \#8.
$$\eqalign{
    w&=(t^3+1)^{100},\cr
    y&=e^{-4t},\cr}\qquad
  \eqalign{
    w'&=100(t^3+1)^{99}\cdot 3t^2=300 (t^3+1)^{99}\cdot t^2\cr
    y'&=-4e^{-4t}\cr}$$


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