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1(a) y = x4-3x2 +2√x+6
dy/dx =4x3-6x+x-1/2
(b) y = 3/3√2x
= 3/ (2x) 1/3
=3(2x) -1/3
dy/dx = -1(2x) -4/3
= -2x -4/3
(c) h(t) = 2t2-3t/5t+1
dt/dh= {(5t+1)(4t-3)-( 2t2-3t)*5}/(5t+1)2
= {20t2-4t-15t-3-(2t2-3t)*5}/25t2+10t+1
= (20t2-11t-3-10t2+15t)/ 25t2+10t+1
=10t2+4t-3/ 25t2+10t+1
2. g(x) = (1-3/x2)(x3-x+2)
dx/dg = (x3-x+2)(-3x-2)+( 1-3/x2)( 3x2-1)
= -3x-6 +6x3-6x-2+3x2-1-9-3x-2
= -3x-6 +6x4-3x-2+3x2-10
3. y = 2/3 x3+ 2 x2-6x+7
slope= Limit ∆x 0 (∆y/∆x)
= 2 x2+4x-6
therefore, dy/dx = 2 x2+4x-6
4(a) P= 50√x-0.5x-500
If x=1000 then,
P = 50 √1000 -0.5 (1000) -500
= 581.13
(b) if x= 800;
= 50√800-0.5*800-500
= 514.21
So, Loss would be 66.92
(c) if x=4000 then
= 50√4000-0.5*4000-500
= 662.27
The marginal profit will be 662.27 at 4000
6. y = (1-x)8
dy/dx = 8(1-x) 7
d2y/d2x=56(1-x)6
d3y/d3x = 336(1-x)5
7 (a) rate of change is (t+1750)/{50(t+2)}
(b) After 1 day the percentage will be
= (1+1750)/{50(1+2)}
= 1751/150
= 11.63%
(c) P'(10) = 1760/600
= 2.93%
If employee works for 10 days the defective percentage will be 2.93%
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