Question

Derivatives

When do you know that you have to use General power rule (chain rule) instead of just the exponent rule?

Answer 1

You are technically always using the general power rule, check this out:

\[f(g(x))=x^3\] Let's say we have this, so then we can break it up as:

\[f(x)=x^3\]\[g(x)=x\]
See for yourself that \(f(g(x))=x^3\) to make sure. (weird I know, but follow along it's insightful!)

So now to use the chain rule we can write (perhaps a little strangely)

\[f'(g(x))g'(x) = 3(g(x))^2 g'(x) = 3x^2 * 1 = 3x^2\]

So in a way you can always sneak in the general rule to the power rule. Perhaps you should try a more complicated one like:

\[f(g(x)) = (x^3)^5\] and see that you get the same answer as if you had just done \[h(x)=x^{15}\]

Answer 2

Does that^ make sense @jaeuni ?

Answer 3

I'm working on trying to understand

Answer 4

The top part I worked out but I'm confused how you translated that directly into the chain rule just because it differs from the examples that I've been looking at

Answer 5

my version is slightly shorter. So, let x=4t+6 ok.

If you are taking the derivative wrt x of f(x)=x^4 what do you get?

Answer 6

well you get \[4x^3*x'\] right?

Answer 7

Isn't the derivative of x^4
4x^3?

Answer 8

well, normally, x' just equals 1. cause dx(x)=1 yea? So it is redundant in the power rule to write x' but it is always there

Answer 9

so once you get a little further, you'll see that the power rule is a special case of the chain rule where x'=1

Answer 10

Sorry I really don't get why there was another x multiplied in

In class all we saw was that the power rule was

and that was it

Answer 11

but in my