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Problem Corner | Solution 275 | by Benjamin Bloch, Ph.D.
1. y’ = (5x - 6)(-3) + (-3x + 4)5 = -15x + 18 - 15x + 20 = -30x + 38. Check
by first expanding the product y = -15x2 +38x -24, and then applying DD1,
becomes y’ = -30x + 38
2. First write 1/x3 as x-3 so that y = (4x2 - 1)x-3, then follow the procedure. Thus
y’ = (4x2 - 1)(-3)x-4 + 8x(x-3) = -12x-2 + 3x-4 +8x-2 = -4x-2 + 3x-4 We can
check our result by first expanding y = (4x2 - 1)x-3= 4x-1 - x-3 so that y’ =
-4x-2 +3x-4
3. First write y = 1/(2x - 3)(x) = (2x - 3)-1(x)-1 then follow the procedure. Thus y’
= (2x - 3)-1(-)(x)-2 + (x)-1(-)2(2x - 3)-2 = -x-2(2x - 3)-1 -2x-1(2x - 3)-2
4. First cast the equation into the proper form by squaring both sides. Then y =
(x2 + 7)2 Then y’ = 2(x2 + 7)2x = 4x3 + 28x. We can check our result by first
expanding y = (x2 + 7)2 = x4 + 14x2 + 49, so that y’ = 4x3 + 28x.
2
February 2011
Professional Surveyor Magazine | www.profsurv.com
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