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How do I use the chain rule to determine the following?

$$\frac{d}{dx}sin(x+c)$$

$$\frac{\partial}{\partial x}sin(x+c)$$

YinWai Tse
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  • Any thoughts? What difference do you see between these two expressions? – lulu Oct 04 '17 at 13:17
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    I see that one is a normal derivative, whereas the other is a partial derivative. But I don't know how to apply the chain rule to solve them. Is there a particular rule (version of the chain rule) I should need to solve this? – YinWai Tse Oct 04 '17 at 13:18
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    As this is a function of one variable, what's the difference between an ordinary derivative and a partial? What version of the chain rule do you know? It's hard to imagine a more basic example of the chain rule than this. – lulu Oct 04 '17 at 13:20
  • In both cases, the result will be the same because $c$ is always considered as a constant. – pitchounet Oct 04 '17 at 13:22
  • May be worth remarking that the chain rule is not even needed here. You can expand $\sin(x+c)$ using the additional formula for $\sin$ and then differentiate directly. Otherwise, the standard chain rule works perfectly: $\frac d{dx} f(g(x))=f'(g(x))\times g'(x)$. – lulu Oct 04 '17 at 13:24
  • Oh, well I've just realised that the results are the same since it is a function of only one variable (I was supposed to make an example where this wasn't the case, but forgot to add a second variable at least). There are 2 versions of the chain rule I know of. The first states "If $y=[f(x)]^n$, then $\frac{dy}{dx}=n[f(x)]^{n-1}f'(x)$ ". The other, I've only seen applications of, but I don't know the actual rule. An example of it being applied is "$\frac{\partial f(x+ct)}{\partial(x+ct)}=\frac{\partial f}{\partial(x+ct)}\frac{\partial(x+ct)}{\partial t}=f' \cdot c$" – YinWai Tse Oct 04 '17 at 13:27
  • Well, look up the Chain Rule. Definitely worth learning the general rule, as opposed to a few special cases of it. – lulu Oct 04 '17 at 13:29
  • What are the other variables involved in your second expression – Anil Kumar Pandey Oct 04 '17 at 13:32

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