How do you find the derivative of $f(x)= 2/(x+1)^2$ ?

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Answer 1 · 2016-07-07T13:01:48

$-4/(x+1)^3$

$u(x) = 2$
$v(x) = (x+1)^2$

$f(x) = (u(x))/(v(x))$

$((u(x))/(v(x)))^' = (u'(x).v(x) - u(x).v'(x))/(v(x)^2$

$f'(x) = (0.(x+1)^2 - 2.(2x+2))/(x+1)^4$

$f'(x) = -(4(x+1))/(x+1)^4 = -(4cancel((x+1)))/cancel((x+1))^4 = -4/(x+1)^3$

Answer 2 · 2016-07-08T18:54:16

$-4/((x+1)^3)$

An alternative view is the following

$f(x) = 2/((x+1)^2) = 2(x+1)^-2$

then we can use the chain rule and power rule

$f'(x) = -4(x+1)^-3 = -4/((x+1)^3)$

How do you find the derivative of f(x)= 2/(x+1)^2f(x)= 2/(x+1)^2?