1. ## Derivative questions?

I'm really not understanding. I thought 4/3 and pi were constants. So I figured the answer was 2ra+r^2
But that is wrong.
Why?

2. 4/3 and pi are constants. They are numbers, connected to the variable r by multiplication.

$\displaystyle V(r) = \frac{4}{3}\pi r^2 a$

$\displaystyle V(r) = \frac{4\pi r^2 a}{3}$

Now we just differentiate the numerator, ignoring the denominator and then simplify afterwards.

3. Well, I just took out 4/3 and pi because I figured since they were constants, their derivatives were nothing? But it seems like you did something different.

Then I thought f = r^2 and g = a
and then f' = 2r and g' = 1

And since they're being multiplied I used the product rule. f'g + fg'

Which gave me (2r)(a) + (r^2)(1)

But that is apparently incorrect? What am I doing wrong?

4. No. Differentiation is linear.

The product rule is only applied if the same variable is being multiplied. Eg: x^2 * cos(x) you would need to use the product rule because it involves the same variable twice.

Consider the following:

$\displaystyle f(x) = 2x^2 + 3$

$\displaystyle f'(x) = 4x + 0$

Correct?
The derivatives are only 0 if they are seperated by a plus or minus. Otherwise, think of the whole thing as a group.

5. Oh, then I never learned how to do this.
But couldn't you take 4/3 and pi out still? Because they are constants regardless?
So then how do you find the derivative with different variables?

6. 1. You can't take out 4/3 and pi. They are constants, but are apart of r^2. I don't see how you cannot understand this yet you know the product rule...

Example:

$\displaystyle f(x)=\frac{7}{9}\pi r^2$

$\displaystyle f'(x)=2 \cdot \frac{7}{9} \cdot \pi r$

See how they still stay with the variable r and don't get taken out?

2. With different variables, think of them as another constant, like pi.

7. I'm sorry, but I never learned this. I'm not quite following.

[(4)(pi)(r^2)(a)]/3

Would become [(8)(pi)(r)(a)]/3

Then what? I really don't understand why you keep the numbers. And I don't know what to do next because everything I thought is wrong apparently.

If I were to think of them as constants, in my mind and from what I've learned and thought this whole time, you would get rid of them.

8. You have got the derivative correct!

You keep them because differentiation is linear.

The constants are connected by the variable by a multiplication or division, and so they are kept together. Constants only get taken away in differentiation if they are NOT connected to the variable you are differentiating with.