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physicskid

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- #1

physicskid

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- #2

Gza

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f(x) = x^2, you take the derivative of the "outside" (x^2) times the derivative of the "inside" (x) yielding f(x) = 2x*(1) . You multiply by one since the chain rule told you to multiply by the derivative of the inside function. I'm not sure if I answered your question, but you'd be better off taking a peek at a good calc textbook (try Stewarts), and working a few problems until you get it. It's really a skill you need to practice to understand how the rules apply.

- #3

rcg

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Although I believe that the power rule is more a derivation of first principles as opposed to application of the chain rule, you can use chain if you want to...but it's so much easier to use anx^n-1 (Power rule).

- #4

Nenad

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Its wery easy, there's no messing around. The power rule is basically the chain rule, but simpler and for easier derivatives.

Lets say we have a function

[tex] u(x) = kx^n [/tex]

[tex] \frac {d}{dx} u(x) = \frac {d(kx^n)}{dx} [/tex]

[tex] \frac {d}{dx} u(x) = (nk)x^{n-1} [/tex]

Now the chain rule. Let's say we have a function:

[tex] f(x) = (u(x))^n [/tex]

[tex] \frac {d}{dx} f(x) = \frac {d((u(x))^n)}{dx} [/tex]

[tex] \frac {d}{dx} f(x)= n(u(x))^n) \frac {d(u(x))}{dx} [/tex]

its a simple set of rules, the best way to get used to them is to practice different examples. Sorry if my notation at the end is a little funky, the latex notation is hard to work with.

Lets say we have a function

[tex] u(x) = kx^n [/tex]

[tex] \frac {d}{dx} u(x) = \frac {d(kx^n)}{dx} [/tex]

[tex] \frac {d}{dx} u(x) = (nk)x^{n-1} [/tex]

Now the chain rule. Let's say we have a function:

[tex] f(x) = (u(x))^n [/tex]

[tex] \frac {d}{dx} f(x) = \frac {d((u(x))^n)}{dx} [/tex]

[tex] \frac {d}{dx} f(x)= n(u(x))^n) \frac {d(u(x))}{dx} [/tex]

its a simple set of rules, the best way to get used to them is to practice different examples. Sorry if my notation at the end is a little funky, the latex notation is hard to work with.

Last edited:

- #5

mathwonk

Science Advisor

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the power rule is the special case where the outer function is a power (g(x))^n,

i.e. here u = g(x) is anything, but f(u) = u^n.

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