Expressing as a composite function to begin differentiating with the chain rule

Question

I'm trying to differentiate $y=(x+1)(x+2)^2(x+3)^3$ using the chain rule, but I am having trouble writing it as a composite function. Any help would be great as I can finish the problem on my own once I just get started in the right direction. Thanks.

Answer 1

Think of it first as a product of $x+1$ and $(x+2)^2(x+3)^3$; differentiating that is going to give you

$$(x+1)\Big((x+2)^2(x+3)^3\Big)'+1\cdot(x+2)^2(x+3)^3\;.$$

Now you have to differentiate $(x+2)^2(x+3)^3$; that’s another product, of $(x+2)^2$ and $(x+3)^3$. It’s only when you differentiate the powers $(x+2)^2$ and $(x+3)^3$ that you really have to invoke the chain rule.

Answer 2

The chain rule alone won't cut it here. You need the rule for differentiating products, i.e. $(f\cdot g)' = f'g + fg'$. Then, first set

$$f(x)=(x+1)(x+2)^2\\ g(x)=(x+3)^3$$

and you can use the chain rule to calculate $g'$, while for $f'$, you need the product rule again.

Answer 3

When you have products such as $$y=(x+1)(x+2)^2(x+3)^3$$ logarithmic differentiation makes life simpler $$\log(y)=\log(x+1)+2\log(x+2)+3\log(x+3)$$ Differentiating $$\frac{y'}y=\frac 1{x+1}+\frac 2{x+2}+\frac 3{x+1}=\frac {6 x^2+22 x+18 }{(x+1) (x+2) (x+3)}$$ Multiply both sides by $y$ to get $$y'=\frac {6 x^2+22 x+18 }{(x+1) (x+2) (x+3)}(x+1)(x+2)^2(x+3)^3=(6 x^2+22 x+18)(x+2)(x+3)^2$$