# Differentiate. G(x)= (6-(1/x)) / (x-2)

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### 3 Answers

G(x) = [(6-(1/x)]/(x-2)

Let G(x) = u/ v such that:

u= 6 - 1/x ==> u' = 1/x^2

v= (x-2) ==> v' = 1

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

= (1/x^2)*(x-2) - (6- 1/x)]/(x-2)^2

= (x-2) - (6x^2 - x)/x^2 (x-2)^2

= (x-2-6x^2 +x)/x^2 (x-2)^2

= (-6x^2 + 2x)/x^(x-2)^2

= -2x(3x-1)/x^2(x-2)^2

** = -2(3x-1)/x(x-2)^2**

To differentiate G(x) =(6-(1/x)/(x-2).

Solution:

G(x) = (6-(1/x))/(x-2) = (6x-1)/(x(x-2) = (6x-1)/(x^2-2x)

We use {u(x)/v(x)} = {u'(x)v(x)-u(x)v'(x)}/(v(x))^2.

In this case, u(x) = (6x-1). So u'(x) = (6x-1)' = 6.

v(x) = (x^2-2x). Therefore v'(x) = (x^2-2x)' = 2x-2.

So G'(x) = {(6x-1)'(x^2-2x)-(6x-1)(x^2-2x)'}/(x^2-2x)^2

={6(x^2-2x)-(6x-1)*(2x-2))}/(x^2-2x)^2

= {6x^2-12x -12x^2+14x-2}/(x^2-2x)^2

={-6x^2+2x-2}/{x(x+2)}^2

= -2(3x^2-x+1)/{x(x-2)}^2.

Therefore G'(x) = -2(3x^2-x+1)/{x(x-2)}^2

We'll note G(x) = y

dy/dx = d/dx {[6-(1/x)] / (x-2)}

dy/dx = d/dx [6/(x-2)] - d/dx [1/x(x-2)]

d/dx [6/(x-2)] = [(x-2)*d/dx(6) - 6*d/dx(x-2)]/(x-2)^2

d/dx [6/(x-2)] = [0*(x-2) - 6*1]/(x-2)^2

d/dx [6/(x-2)] = - 6/(x-2)^2 (1)

d/dx [1/x(x-2)] = d/dx [1/(x^2-2x)]

d/dx [1/(x^2-2x)] = [(x^2-2x)*d/dx(1) - 1*d/dx(x^2-2x)]/x^2*(x-2)^2

d/dx [1/(x^2-2x)] = -(2x-2)/x^2*(x-2)^2 (2)

dy/dx = (1) - (2)

dy/dx = - 6/(x-2)^2 + (2x-2)/x^2*(x-2)^2

**dy/dx = (2x - 2 - 6x^2)/x^2*(x-2)^2**

**dG/dx = 2(x-1-3x^2)/x^2*(x-2)^2**