Differentiation of a function in a domain

[paech]

Homework Statement

Find the derivatives at an arbitrary point $x$ in the domain of the following functions $f_i: D_i → ℝ$, where for $1 ≤ i ≤ 6$ the domain $D_i$ is the maximal subset of $ℝ$ on which the mapping is defined - you don't have to determine the domains.

Homework Equations

a) $f_1 (a) = (1-\sin3a)^5$

There are more functions for other parts of the question, but I just need help with understanding the problem.

The Attempt at a Solution

Ok, so I'm having trouble understanding what the question is really saying. From what I gather, it's saying that the domain it's listed is the set of real numbers in which the derivative will be defined. The domain it's listing has put me off, usually I would just substitute the arbitrary point $x$ into the function like so:$$f'(x) = (1-\sin3x)^5$$ and then apply the chain rule to find the derivative. I'm not sure what to do to find the derivative in the interval given.

Sorry if I haven't provided much of an attempt at a solution, I've looked over all my material and couldn't come up with anything.

 

[HallsofIvy]

Homework Statement

Find the derivatives at an arbitrary point $x$ in the domain of the following functions $f_i: D_i → ℝ$, where for $1 ≤ i ≤ 6$ the domain $D_i$ is the maximal subset of $ℝ$ on which the mapping is defined - you don't have to determine the domains.

Homework Equations

a) $f_1 (a) = (1-\sin3a)^5$

There are more functions for other parts of the question, but I just need help with understanding the problem.

The Attempt at a Solution

Ok, so I'm having trouble understanding what the question is really saying. From what I gather, it's saying that the domain it's listed is the set of real numbers in which the derivative will be defined. The domain it's listing has put me off, usually I would just substitute the arbitrary point $x$ into the function like so:$$f'(x) = (1-\sin3x)^5$$

Was this a typo? $f(x)= (1-\sin(3x))^5$, not its derivative. And you do not substitute the value of x first and then differentiate! The value of a function is a constant and the derivative of a constant is always 0.

and then apply the chain rule to find the derivative. I'm not sure what to do to find the derivative in the interval given.

Sorry if I haven't provided much of an attempt at a solution, I've looked over all my material and couldn't come up with anything.

You are reading too much into this problem. All it is asking you to do is find the derivative- in its domain, which is nothing new. A function isn't defined outside its domain so this is just what you have been doing ever since you started differentiation. And the problem specifically says "you don't have to find the domains". Just differentiate!

 

[vela]

Homework Statement

Find the derivatives at an arbitrary point $x$ in the domain of the following functions $f_i: D_i → ℝ$, where for $1 ≤ i ≤ 6$ the domain $D_i$ is the maximal subset of $ℝ$ on which the mapping is defined - you don't have to determine the domains.

Homework Equations

a) $f_1 (a) = (1-\sin3a)^5$

There are more functions for other parts of the question, but I just need help with understanding the problem.

The Attempt at a Solution

Ok, so I'm having trouble understanding what the question is really saying. From what I gather, it's saying that the domain it's listed is the set of real numbers in which the derivative will be defined.

You made a slight error. D_i is the domain of f_i, not f'_i, though the domains of the two could very well be the same.

The problem statement is just explicitly saying what you've likely been assuming all the time. When you define a function, you have to specify its domain, but it's common simply to say things like "Let f(x)=x+1" with the understanding that the domain is the set of all real numbers for which the function is defined. So like Halls said, the problem is really just asking you to find the derivatives like you normally do.

 

[paech]

Ah right, thanks. I had a feeling that I was misinterpreting the problem and that I just needed to differentiate it as I normally would.