find the derivative using quotient rule y = (x + 1)^2(x2 + 1)^-3 step by step pls

let f(x)=g(x)/h(x)
The quotient rule states that:
f'(x)=[g'(x)h(x)-g(x)h'(x)]/[h(x)]²
From the expression given:
y = (x + 1)^2(x2 + 1)^-3
y=(x+1)²/(x²+1)³
where:
g(x)=(x+1)²
g'(x)=2(x+1)

h(x)=(x²+1)³
h'(x)=6x(x+1)²
hence to get the derivative of y we substitute in the formula:
y'=[2(x+1)(x²+1)³-6x(x+1)²(x+1)²]/[(x²+1)^6]
simplifying the above we get:
y[2(x+1)(x²+1)³-6x(x+1)^4]/[(x²+1)^6]

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ejhuff ejhuff · Answer 2 · 2019-01-06T03:13:43+01:00

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Find the derivative using quotient rule [tex]y\ =\ \frac{\displaystyle (x + 1)^}{\displaystyle 2(x^2 + 1)^3}[/tex].

Previous answer due to CastleRook has an error computing h'(x).

Quotient rule:[tex]f(x)\ =\ \frac{\displaystyle g(x)}{\displaystyle h(x)}\\f'(x)\ =\ \frac{\displaystyle g'(x)h(x)\,-\,g(x)h'(x)}{\displaystyle (h(x))^2}[/tex]

In the problem,[tex]f(x)\ =\ \frac{\displaystyle (x + 1)^2}{\displaystyle (x^2 + 1)^3}\\g(x)\ =\ (x+1)^2\\g'(x)\ =\ 2(x+1)\\h(x)\ =\ (x^2+1)^3\\h'(x)\ =\ 3(x^2+1)^22x\ =\ 6x(x^2+1)^2[/tex]Substituting,

[tex]f'(x)\ =\ \frac{\displaystyle g'(x)h(x)\,-\,g(x)h'(x)}{\displaystyle (h(x))^2}\\\\\ \ =\ \frac{\displaystyle 2(x+1)(x^2+1)^3-(x+1)^26x(x^2+1)^2}{\displaystyle (x^2+1)^6}\\\\\ \ =\ -\frac{\displaystyle 2 (x+1) \left(2 x^2+3 x-1\right)}{\displaystyle \left(x^2+1\right)^4}[/tex]