Implicit Differentiation

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Prepared for Gina. L

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1) find dy/dx (or y') for xy2 = 3 

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Solution

d/dx(xy2) = d/dx(3) 

x(2y)(dy/dx) + y2 = 0

2xy(dy/dx) = -y2

dy/dx = -y2/(2xy)

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2) find dz/dx (or z') for x2yz = x (z, y are both a function of x) 

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Solution

d/dx(x2yz) = d/dx(x)

d/dx((x2)(yz)) = 1

(x2)d/dx(yz) + (yz) d/dx(x2) = 1

(x2)(y dz/dx + z dy/dx) + (yz)(2x) = 1

(x2)(y dz/dx + z dy/dx) = 1 - (2xyz)

(y dz/dx + z dy/dx)  = (1 - (2xyz)) /(x2)

y dz/dx = ((1 - (2xyz)) /(x2)) - (z dy/dx)

dz/dx = (((1 - (2xyz)) /(x2)) - (z dy/dx)) / y

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example 

d/dx ABC can be done by grouping them.

d/dx ABC

= d/dx (A)(BC)

= A (BC)' + (BC) A'

= A (BC' + CB') + BCA'

= ABC' + ACB' + BCA'

= ABC' + AB'C + A'BC@

 

3) Given a circle equations x2 + (y+ 1)2 = 16

find the following...

3.1 Radius of the circle

3.2 Centre of the circle

3.3 The equation of the tangent pass through circle at point (0, 3).

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Solution

3.1

r = 16 = 4

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3.2

Centre point is (0, -1)

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Notice: If we have another circle equation as (x - 2)2 + (y- 3)2 = 16

than the centre point is (2, 3)

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3.3

d/dx[x2 + (y+ 1)2] = d/dx (16)

2x + 2(y+1)(dy/dx) = 0

2(y+1)(dy/dx) = -2x

dy/dx = -2x/(2(y+1))

dy/dx = -x/(y+1)

sub x =0, y =3 into dy/dx

dy/dx = -0/(3 + 1)

          = 0

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Now, we have the slope of the tangent,

m (slope) = dy/dx = 0

y = mx + b

y = (0)x + b

y = b

b = 3

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Answer: the tangent equation is y = 3

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written by : Kevin T (Mar 8, 2010)