What is the derivative of f(x) = (x^2-4)/(x-2) using the quotient rule?

Question

(in simplest form)

NOTE: If you simplify f(x) = (x^2-4)/(x-2) first, and THEN take the derivative, the first derivative is 1

Answer 1

f (x) = ( x ² - 4 ) / ( x - 2 )
f `(x) = [ ( x - 2 )( 2 x ) - ( x ² - 4 )(1) ] / (x - 2) ²
f `(x) = [ x ² - 4 x + 4 ] / ( x - 2 ) ²
f `(x) = ( x - 2 ) ² / ( x - 2 ) ²
f `(x) = 1

Answer 2

The quotient rule states that:

d(u/v)/dx = (v*du/dx - u*dv/dx) / v^2

In this case, u = x^2 - 4, and v = x - 2. Deriving u and v will give you:
du/dx = 2x
dv/dx = 1

Plugging these values into the above equation for the quotient rule gives you:

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

u, incidentally, factorises to (x - 2)(x + 2), and so you can factorise the top of the fraction, and reduce the whole fraction by dividing both numerator and denominator by (x - 2):

dy/dx = ((x - 2)2x - (x^2 - 4)) / (x - 2)^2
= (x - 2)(2x - (x + 2)) / (x - 2)(x - 2)
= (2x - x - 2) / (x - 2)
= (x - 2) / (x - 2)
= 1

Answer 3

the quotient rule is g"(x) f(x) - f ''(x) g(x) / (g (x) ) ^ 2...

take note that the rule is not commutative, ( " ) means derivative of... Try to apply the rule to get the answer.

Answer 4

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

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

Both ways are correct..

Answer 5

(x-2)(2x)-(1)(x^2-4)/(x-2)^2

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

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

=1