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I want to find the derivative of the function: $h(x) = x^2(x+1)$

We know that the correct derivative is $h'(x) = 3x^2+2x$

I want to know if chain rule can as well be applied here: $h′(x)=f′(g(x))∗g′(x)$

Can we say the following?

We have the inner function $g(x)$, which adds one to the argument:

$g(x) = x+1$

And we have the outer function $f(x)$, which multiplies the argument by $x^2$:

$f(x) = x^2*x$ (or better write $f(y) = x^2*y$ to avoid confusion?).

But if I proceed and say $f'(x)=3x^2$ and $g'(x) = 1$, then $h′(x)=f′(g(x))∗g′(x) = 3(x+1)^2*1=3x^2+6x+3 \neq 3x^2+2x$

Please let me know where I am wrong and if chain rule is applicable here at all. I know that in this situation product rule is better to be used here. But if chain rule cannot be applied here, please explain where the logic breaks.

PS. I have seen a similar question here Finding derivative using product and chain rule and it implies that chain rule may be used.

Антон Бугаев
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  • This has nothing to do with the chain rule, since $h(x)=f(x)g(x), $ a product (!). – Fred Mar 24 '22 at 10:02
  • You are confusing composition of functions with products. – Kavi Rama Murthy Mar 24 '22 at 10:03
  • But why can't we say that product is a special case of composition? Can't we say that a product is a function, which multiplies some argument (which can as well be a function) by some value? – Антон Бугаев Mar 24 '22 at 10:23
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    Of course we can! Let $p\colon\mathbb R^2\to\mathbb R$ defined by $p(x,y)=x\cdot y$. Then $$\nabla p(x,y)= \begin{pmatrix} y\x\end{pmatrix}.$$ For real-valued functions $f$ and $g$ we get from the cain rule $$p(f, g)= \langle \begin{pmatrix} g\f\end{pmatrix},\begin{pmatrix} f'\g'\end{pmatrix}\rangle=gf'+fg'.$$ – Michael Hoppe Mar 24 '22 at 19:34

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The 'trick' is that you have to apply the multivariate version of the chain rule. When function is a composition of a bivariate function with arguments that each are monovariate functions :

$${[f\circ (g,h)]'(x) = \bigl[g'\cdot f^{(1,0)}\circ(g,h)+h'\cdot f^{(0,1)}\circ(g,h)\bigr](x)}$$

Now let $h(x) = f(x, x+1)$ where $f(x,y)=x^2y$. $$h(x)= x^2~(x+1)$$

Then we have the partial derivatives $f^{(1,0)}(x,y)=2xy$ and $f^{(0,1)}(x,y)=x^2$ .

$${f^{(1,0)}(x,x+1)=2x~(x+1)\\f^{(0,1)}(x,x+1)=x^2}$$

The above chain rule gives us: $$\begin{align}h'(x) &= (x)' f^{(1,0)}(x,x+1)+ (x+1)' f^{(0,1)}(x,x+1)\\&= 2x~(x+1)+x^2\\&=3x^2+2x\end{align}$$

Graham Kemp
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