Find the 2009-th derivative of the function f(x)=1/(x^2-4). Calculate for x=0.

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neela | High School Teacher | (Level 3) Valedictorian

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Let y = 1/(x^2-4)

dy/dx = y1 = [-1/(x^2-4)^2]*2x. Or

y1*(x^2-4) = -2x. Or

(x^2-4)*y1 = -2x. ....(1)

(y1)x=0 = -2*0/(0-4) = 0. Or y1 = 0.

Differentiating (1)  again,  2xy1+x^2*y2 = 0 from which [y2 at x=0] = 0. Or y2 = 0.

Takiing nth derivative of (1), by Leibnitz rule  on both sides, we get:

Dn{ y1(x^2-4) y} =  Dn(-2x) = 0  Or

Dn(y1)*(x^2-4)+ nC1*Dn-1Yn-1* D(x^2-4)+nC2*Dn-2*Yn*D2(x^2-1)+ nC3*Dn-3y1*0+ 0...  = 0

= (yn+1)(x^2-4)+n*yn*(2x)+n(n-1)/2] yn-1(2)

 At x = 0, ( yn+1)(-4)+n(n-1)yn-2 = 0

Therefore , yn+1 =  [n(n-1)yn-1]/4.

When n= 2, at x=0,  y3 = (2*1*y1)/4 = {2*1 *0/4} = 0 .

n = 3, at x=0, y4 =  3*2*y2/4  = 0, as y2 = 0.

Continuing like this , we get, we get at x=0, y3 =y5=y7= ...y2n+1 = 0. Or

So ,at x= 0, the 2009th derivative y2009  = 0

Top Answer

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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First, we notice that the denominator of the function is a difference of squares, so we'll re-write it as:

 f(x)=1/(x^2-4) = 1/(x-2)(x+2)

Now, we'll write the function as a sum or difference of 2 irreducible ratios.

1/(x-2)(x+2) = A/(x-2) + B/(x+2)

We'll calculate the common denominator, to the right side and we'll multiply each ratio with the corresponding amount.

1 = Ax+Bx + 2A - 2B

After factorization we'll get:

1 = x(A+B) + 2(A-B)

The expression from the left side could be written as:

1 = 0*x + 1

So, for the identity to hold, the both expressions from both sides have to correspond.

A+B = 0

2(A-B) = 1

A-B = 1/2, but A=-B

2A = 1/2

A=1/4

B=-1/4

1/(x-2)(x+2) = 1/4(x-2) - 1/4(x+2)

Now, we'll calculate the first derivative:

f'(x) = [1/4(x-2) - 1/4(x+2)]'

f'(x) = 4/16(x+2)^2 - 4/16(x-2)^2

f'(x) = 1/4(x+2)^2 - 1/4(x-2)^2

We'll calculate the second derivative:

f''(x) = 8(x-2)/16(x-2)^4 - 8(x+2)/16(x+2)^4

After reducing the terms:

f''(x) = 2/4(x-2)^3 - 2/4(x+2)^3

And to establish a final form, we'll calculate the third derivative:

f'''(x) = 2*3*4(x+2)^2/16(x+2)^6 - 2*3*4(x-2)^2/16(x-2)^6

After reducing the terms:

f'''(x) = 2*3/4(x+2)^4 - 2*3/4(x-2)^4

It is obvious that:

f(x)^(2009) = 2009!/4(x+2)^2010 - 2009!/4(x-2)^2010

Now, we'll calculate for x=0

We'll factorize:

f(0)^(2009) = 2009!/4(1/2^2010 - 1/2^2010)

f(0)^(2009) = 0

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