# If p(x)= f(x^3), find P'(1)

#### Willian93

1. The problem statement, all variables and given/known data

if p(x)= f(x^3), find P'(1)

2. Relevant equations

3. The attempt at a solution

can i do the derivative of x^3 is 2x^2, then substitute to get f(2x^2)?, then substitute 1 for x?

#### cepheid

Staff Emeritus
Gold Member
Re: Derivative

1. The problem statement, all variables and given/known data

if p(x)= f(x^3), find P'(1)
Are both p's supposed to have the same case here (i.e. are they the same symbol representing the same function?) If not, then this question doesn't make much sense.

can i do the derivative of x^3 is 2x^2, then substitute to get f(2x^2)?, then substitute 1 for x?
No, not quite. If I understand the problem right, then what you have is a composite function, and you need to use the chain rule, and you need to know what the functional form of f(x) is. Is f(x) given?

One way of looking at it: p(x) = f(g(x)) where g(x) = x3. Hence, a composite function.

Another equivalent way to look at it: f(x3) means, "pass x3 as an argument into the function f." In other words, p(x) is the function you get when you pass u = x3 as an argument into the function f(u), where I have used a different symbol for the argument passed into f in order to be more explicit.

Edit: Those two ways of looking at it are not really different at all, since in the latter case, I just defined a new variable u = g(x) = x3. But I hope that the wording/explanation helps you to understand what is going on.

#### Char. Limit

##### PF SAS Commando
Gold Member
Re: Derivative

Also, the derivative of x^3 is not 2x^2.

#### HallsofIvy

Re: Derivative

1. The problem statement, all variables and given/known data

if p(x)= f(x^3), find P'(1)

2. Relevant equations

3. The attempt at a solution

can i do the derivative of x^3 is 2x^2, then substitute to get f(2x^2)?, then substitute 1 for x?
No. The chain rule says that the derivative of f(g(x)) is f'(g(x))g'(x). (And, as Cepheid said, the derivative of x^3 is NOT "2x^2".)

#### cepheid

Staff Emeritus
Gold Member
(And, as Cepheid said, the derivative of x^3 is NOT "2x^2".)
Actually, Char. Limit said that. It was a good catch...

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