- #1

- 403

- 106

- Summary:
- Applying the chain rule in a way that is not often encountered.

I want to take the derivative of a composite function that looks like

$$f( g(x), h(x) ).$$

I know from Wolfram that the answer is

$$\frac{ df( g(x), h(x) ) }{ dx } = \frac{ dg(x) }{ dx }\frac{ df( g(x), h(x) ) }{ dg(x) } + \frac{ dh(x) }{ dx }\frac{ df( g(x), h(x) ) }{ dh(x) }.$$

We can generalize the result to get

$$\frac{ df( g_{1}(x), g_{2}(x), ..., g_{n}(x) ) }{ dx } = \sum_{i=1}^{n}{ \frac{ dg_{i}(x) }{ dx }\frac{ df( g_{1}(x), g_{2}(x), ..., g_{n}(x) ) }{ dg_{i}(x) } }.$$

However, I am struggling to understand why. Where does the summation come from? It's like the product rule was applied somewhere, but I'm not seeing where.

Can anyone help me to understand this?

Thanks,

Zap.

$$f( g(x), h(x) ).$$

I know from Wolfram that the answer is

$$\frac{ df( g(x), h(x) ) }{ dx } = \frac{ dg(x) }{ dx }\frac{ df( g(x), h(x) ) }{ dg(x) } + \frac{ dh(x) }{ dx }\frac{ df( g(x), h(x) ) }{ dh(x) }.$$

We can generalize the result to get

$$\frac{ df( g_{1}(x), g_{2}(x), ..., g_{n}(x) ) }{ dx } = \sum_{i=1}^{n}{ \frac{ dg_{i}(x) }{ dx }\frac{ df( g_{1}(x), g_{2}(x), ..., g_{n}(x) ) }{ dg_{i}(x) } }.$$

However, I am struggling to understand why. Where does the summation come from? It's like the product rule was applied somewhere, but I'm not seeing where.

Can anyone help me to understand this?

Thanks,

Zap.