question involving quoitient and product rule

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I have already asked this question..however this time Ill show my work and I need help on finding where I messed up. 5.d) Find dy/dx at the given value of x. y=[(x+1)(x+2)]/[(x-1)(x-2)], x=4

first I did used the product rule: dy/dx= [1(x+2)+(x+1)1]/[1(x-2)+(x-1)1]
dy/dx= [x+2+x+1]/[x-2+x-1]
dy/dx= [2x+3]/[2x-3]

and now I would use the quoitient rule, dy/dx= [2(2x-3)] -[(2x+3)(2)]/(2x-3)^2
dy/dx= (4x-6)-(4x+6)/(2x-3)^2
dy/dx=-12/(2x-3)^2

SUb in 4 into x,: -12/ 25<--this is what I got for my final answer, however it is suppose to be -7/3 according to the text book, so HELPPP
Thank you for the help
 
y=[(x+1)(x+2)]/[(x-1)(x-2)]

expand the numerator and denominator ...

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

now use just the quotient rule ...

dy/dx = [(x^2 - 3x + 2)(2x + 3) - (x^2 + 3x + 2)(2x - 3)]/(x^2 - 3x + 2)^2

now, evaluate the derivative at x = 4.
 
but shouldn't it also work if I used the product rule? because are teacher won't let us expand, she wants us to use the rules
 
yes, but you have to use both rules at the same time ... not one then the other.

to suit your teacher ...

y=[(x+1)(x+2)]/[(x-1)(x-2)]

dy/dx = [(x-1)(x-2){(x+1)(1)+(x+2)(1)}-(x+1)(x+2)*{(x-1)(1)+(x-2)(1)}]/[(x-1)(x-2)]^2

now evaluate at x = 4
 
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