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I am new to calculus, and thought I had my head round the product, quotient and chain rules, but I can't work out how to tackle this:

$$ \frac 1{x(x+1)^2} $$

Apparently, the first step of the solution is

$$ f'(x) = - \frac {(x+1)^2 + 2x(x+1)} { x^2(x+1)^4} $$

but I can't work out how this was arrived at. The denominator of the derivative is the square of the denominator of the function, which suggest use of the quotient rule, but by what steps was the numerator of the derivative arrived at? I expect it is my poor algebra and manipulation techniques letting me down. If anyone can recommend a good tool for improving these (as well as answering my question!) I"d be grateful.

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I'd actually use the chain rule if I were in your shoes. Treat your function as $f(g(x))$, where $f(x)=\dfrac1{x}$ and $g(x)=x(x+1)^2$. Another possibility is to split into partial fractions... – J. M. Aug 5 '12 at 16:05
up vote 5 down vote accepted

Apply the quotient rule with $u=1$, $v=x(x+1)^2$, then

$$\left(\frac 1 v\right)' = \frac{(1)'v-1v'}{v^2} = \frac{0-v'}{v^2}$$

because $(1)'=0$. Now for the derivative of $v$ in the numerator, use product rule.

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Thanks Dario, this was a helpful way to approach it. Here's Thanks Dario, this was a helpful way to approach it. Here's how I tackled the derivative of v, but it seems long-winded - is there a better way to do it?$$ v' = (2x+1)(x+1)+(x^2+x)(1) = 2x^2+3x+1+x^2+x = 3x^2+4x+1 = (3x+1)(x+1) $$ – pocketdora Aug 5 '12 at 19:57
Seems a valid approach, however given that the resulting derivative factors so nicely (we can cancel $(x+1)$), it would be cool to keep track of that factor while deriving. Just use plain product rule, you'll get $$(x(x+1)^2)' = (x+1)^2 + x\cdot 2(x+1) = (x+1+2x)(x+1)=(3x+1)(x+1)$$ directly. – Dario Aug 6 '12 at 8:45

You can use quotient rule, as mentioned, or, as my father always prefers, exponent rule and chain rule. Note


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Let $\frac{1}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$ applying partial fraction.

So, $1=A(x+1)^2+Bx(x+1)+Cx$

Putting x=0 in the identity or comparing the constants, A=1

Putting x=-1 in the identity, C=-1

Comparing the coefficient of $x^2$, 0=A+B=>B=-A=-1.

So, $\frac{1}{x(x+1)^2}=\frac{1}{x}-\frac{1}{x+1} -\frac{1}{(x+1)^2}$

Now, $\frac{d{(x+c)^{-n}}}{dx}=-\frac{n}{(x+c)^{(n+1)}}$ for any positive integer n.

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If you know well the general rule $$\left(\frac{1}{f(x)}\right)'=-\frac{f'(x)}{f(x)^2}$$ you can use here the product rule $\,(h(x)g(x))'=h'(x)g(x)+h(x)g'(x)\,$ : $$f(x):=\frac{1}{x(x+1)^2}=\frac{1}{x}\frac{1}{(x+1)^2}\Longrightarrow f'(x)=-\frac{1}{x^2}\frac{1}{(x+1)^2}-\frac{1}{x}\frac{2}{(x+1)^3}=$$ $$=-\frac{1}{x(x+1)^2}\left[\frac{1}{x}+\frac{2}{x+1}\right]$$ or changing the exponents' signs and using $\,(x^n)'=nx^{n-1}\,\,,\,\,\forall\,n\in\Bbb R\,$ , and the product rule again: $$f(x)=x^{-1}(x+1)^{-2}\Longrightarrow f'(x)=(-1)x^{-2}(x+1)^{-2}+x^{-1}(-2)(x+1)^{-3}=$$ $$=-x^{-1}(x+1)^{-2}\left(x^{-1}+2(x+1)^{-1}\right)$$

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Let $$y=\frac{1}{x(x+1)^2}\implies x(x+1)^2y=1$$ Now differentiate both sides and use product rule for L.H.S. which gives $$y((x+1)^2+2x(x+1))+y'x(x+1)^2=0$$ $$\implies y(x+1)(3x+1)+y'x(x+1)^2=0$$ $$\implies y'=-\frac{y(x+1)(3x+1)}{x(x+1)^2}=-\frac{(3x+1)}{x^2(x+1)^3}$$

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