How do you differentiate $y=(x+3)^2/(x-1)$ ?

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Answer 1 · 2016-05-13T00:38:51

$dy/dx = ((x-5)(x+3))/(x-1)^2$

$y = (x+3)^2/(x-1)$

The quotient rule states that if $y = f(x)/g(x)$

Then $dy/dx = (g(x)*f'(x) - f(x)*g'(x))/(g(x))^2$

Using the quotient rule in this example:
$dy/dx$ = $((x-1)*2(x+3)*1 - (x+3)^2*1) / (x-1)^2$

$dy/dx = (2(x^2+2x-3) - (x^2+6x+9)) / (x-1)^2$

$dy/dx= (x^2-2x-15)/(x-1)^2$

$dy/dx= ((x-5)(x+3))/ (x-1)^2$

How do you differentiate y=(x+3)^2/(x-1)y=(x+3)^2/(x-1)?