Asked by hi
Hey I am working on differentiation/quotient rule for this problem...can someone help me?
x^2-3/x-2.
x^2-3/x-2.
Answers
Answered by
Steve
if y=u/v,
y' = (u'v-uv')/v^2
so, let u = x^2-3, u'=2x
let v = x-2, v'=1
y' = ((2x)(x-2)-(x^2-3)(1))/(x-2)^2
= (2x^2-2x-x^2+3)/(x-2)^2
= (x^2-4x+3)/(x-2)^2
= (x-3)(x-1)/(x-2)^2
or, you can always think of it as a product rule:
y = (x^2-3) * (x-2)^-1
y' = (2x-3)/(x-2) - (x^2-3)*(x-2)^-2
which works out the same
y' = (u'v-uv')/v^2
so, let u = x^2-3, u'=2x
let v = x-2, v'=1
y' = ((2x)(x-2)-(x^2-3)(1))/(x-2)^2
= (2x^2-2x-x^2+3)/(x-2)^2
= (x^2-4x+3)/(x-2)^2
= (x-3)(x-1)/(x-2)^2
or, you can always think of it as a product rule:
y = (x^2-3) * (x-2)^-1
y' = (2x-3)/(x-2) - (x^2-3)*(x-2)^-2
which works out the same
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