f(x) = 2x/ (x-2)^2 find f'(1)

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f(x) = 2x / (x-2)^2

Let f(x) = u/v where:

u = 2x ==> u' = 2

v= (x-2)^2 ==> v' = 2(x-2) = 2x-4

f'(x) = (u'v - uv')/v^2

= [2*(x-2)^2 - 2x(2x-4)]/(X-2)^4

= (2x^2 -8x +8 -4x^2 + 8x)/(x-2)^4

= -2x^2 +8)/(x-2)^4

= -2(x^2 -4)/(x-2)^4

= -2(x-2)(x+2)/(x-2)^4

= -2(x+2)/(x-2)^3

f'(1) = -2(3)/(-1)^3

= -6/-1= 6

==> f'(1) = 6

To calculate f'(1), first we have to calculate f'(x), using the quotient rule:

f'(x) = {(2x)'*(x-2)^2 - (2x)*[(x-2)^2]'}/(x-2)^4

f'(x) = [2*(x-2)^2 - 2*2x*(x-2)] / (x-2)^4

We'll factorize:

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

We'll divide by (x-2):

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

Now, we'll calculate f'(1):

f'(1) = -2(1+2)/(1-2)^3

f'(1) = -6/-1

**f'(1) = 6**

f(x) = 2x/(x-2)^2.

To find f'(1).

Let us have transformation.

t = x-2, sothat dt = dx , and when x= 1, t =1-2 =-1.

2x/(x-2)^2 = 2(t+2)/t^2 = 2t+4/t = g'(t)

Therefore,

f'(1) = g'(-1) = {2/t+4/t^2 , t =-1

= -2/t^2 + 4(-2)/t^3 , as t^n)' = nt^(n-1) = 1-n.(1/t^n+1).

g'(-1) = -2/(-1)^2 + (-8/(-1)^3

= -2/1 -8(-1)

= -2 + 8

= 6.

So f'(1) = 6

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