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Free Math Essay Sample

. Differentiate with respect to x


F(x) =   1/3 x7

F(x) = 1/3x -7

FI(x) = -7/3x-8

fI(x)= -7/3x8

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If f(x) =  g(x)/h(x)

fI(x)= (h(x) gI(x) - g(x)hI(x))/h(x)2

=(x-5) (12x2-4x3) / (x-5)2

= (12x3-60x2-4x3)/ (x2-10x+25)

= (  8x3-60x2) /(x2-10x+25)


F(x) = (2x2-6)3(3x3-3)

F(x) = g(x)h(x)

fI(x)= g(x) hI(x) + h(x)gI(x)

fI(x) = (2x2-6)2(3x3-3)I+((2x2-6)3)I(3x3-3)

= (2x2-6)2(2x2-6)27x2+((2x2-6)3)I(3x3-3)


= (2x2-6)2(2x2-6)27x2+12x (2x2-6)(3x3-3)

=(4x4-24x2+36)(2x2-6)27x2+12x (4x4-24x2+36)(3x2-3)

=(4x4-24x2+36)(54x4-162x2)+ (48x5-288x3+432x)(3x2-3)

=216x8-1296x8+1944x4-648x8+3888-5832x2 +144x7+864x5+1296x3-144x7+864x5-1296x




F(x) = xln(2x)

We know dlnU/dx=UI/U

Also F(x)=g(x)h(x)


= 2x/2x+ln (2x)

Answers to question 2


TR=P x Q,                           TR- Total Revenue

TR = 400Q - Q2/50

Expression for marginal revenue is,

MR = 400 - Q/25

MR = 400 - 10000 / 25


Price elasticity demand when price=100

P = 400 - Q/50

Q/50 = 400 - P

Q=400 x 50 - 50P,

Therefore Gradient =-50

Price elasticity=gradient x P/Q

=-50 x 100/10000

=-1/3 hence inelastic

c) Henry can maximize daily revenue by either increasing the quantity of the commodities. Increase in price may not affect demand for the commodity due to the inelastic demand for the commodity. Increasing the quantity is the only way to maximize profit hence revenue.


Question three answers

a)     A=P(1+R)n






R=I/PT = 0.2682P / P=0.2682


b) A=P(1+R)n


A= P(1+6/400)24


A=1.4295 x 12000


C) 5/100x 1000x 40 = £ 2000

4. Express the first and the second order derivatives of;



fI(x,y) = 12x + 5 dy/dx

fII(x,y) = 12 + d2y/dx2


(b) f(x,y) = 3x5y4

fI(x,y)= 15x4y4+3x54y3dy/dx

fII(x,y)= 60x3y4+15x44y3dy/dx+15x4y3dy/dx+3x512y2d2y/dx2


c) f(x,y)=(10+x2y)3


=3(10+x2y)2 (2xy+ x2dy/dx)

Using f(x) =h(x)g(x)

fI(x,y)= h(x)gI(x)+hI(x)g(x),

fI(x,y) = (10+x2y)2(2y+ 2xdy/dx+2xd2y/dx2)+(2xy+x2dy/dx)





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