Maths M2@(20pts)

6.
xy + y^8 = 2009
(d/dx)(xy + y^8) = (d/dx)(2009)
x(dy/dx) + y + 8y^7(dy/dx) = 0
(x + 8y^7)(dy/dx) = -y
dy/dx = -y/(x + 8y^7)


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7)
dy/dx
= d(lnx^2- 2x)^5/dx
= [d(lnx^2 - 2x)^5/d(lnx^2- 2x)] • [d(lnx^2 - 2x)/dx]
= 5(lnx^2 - 2x)^4 • 2[d(lnx - x)/dx]
= 10(lnx^2 - 2x)^4 • [(1/x) - 1]
= 10(lnx^2 - 2x)^4 (1 - x) / x


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8)
dy/dx
= dcos(2x- 9)^3/dx
= [dcos(2x - 9)^3]/d(2x- 9)^3] • [d(2x - 9)^3/d(2x - 9)] • [d(2x - 9)/dx]
= (-sin(2x - 9)^3•3(2x - 9)^2 • 2
= -6 (2x - 9)^2 sin(2x - 9)^3


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9)
y = 6x / (3x - 8)

dy/dx
= [(3x - 8)•(d6x/dx) - 6x•d(3x - 8)] / (3x - 8)^2
= [6(3x - 8) - 18x] / (3x - 8)^2
= (18x - 48 - 18x) / (3x - 8)^2
= -48 / (3x - 8)^2

d^2y/dx^2
= -48•[d(3x - 8)^-2/dx]
= -48•[d(3x - 8)^-2/d(3x - 8)]•[d(3x - 8)/dx]
= -48•(-2(3x - 8)^-3)•3
= 288 / (3x - 8)^3


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10)
a)
y = [(x^3)/2] - 2x
dy/dx = [(3x^2/2)] - 2

(2, 0) lies on the curve.
Slope of tangent at (2, 0)
= {[3(2)^2]/2} - 2
= 4

Equation of the tangent to the curve at (2, 0) is (point-slope form) :
(y - 0) = 4(x - 2)
y = 4x - 8
4x - y - 8 = 0


b)
y = [(x^3)/2] - 2x
dy/dx = [(3x^2/2)] - 2

(-2, 0) lies on the curve.
Slope of tangent at (-2, 0)
= {[3(-2)^2]/2} - 2
= 4

Slope of normal at (-2, 0)
= -1/4

Equation of the normal to the curve at (-2. 0) is (point-slope form) :
y - 0 = (-1/4)(x + 2)
4y = -x - 2
x + 4y + 2 = 0

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