# Thread: Stuck on a derivative problem! (d/dr)[(4 pi r^3)/3] = 4 pi r^2

1. ## Stuck on a derivative problem! (d/dr)[(4 pi r^3)/3] = 4 pi r^2

My calc class is studying basic derivatives, and I'm stuck on a problem.
The problem is f(r) = (4/3) * pi * r3 The answer is 4pir2

$\dfrac{d}{dr}\, \left[\dfrac{4\pi r^3}{3}\right]$

. . .$=\, \dfrac{4\pi}{3}\, \cdot\, \dfrac{d}{dr}\left[r^3\right]$

. . .$=\, \dfrac{4\pi\, \cdot\, 3r^2}{3}$

. . .$=\, 4\pi r^2$

I understand that the power rule calls for the exponent of 3 to be brought to the front and then the exponent is decreased by one. I also understand that the 3 in the denominator cancels out with the 3 in the numerator.
What I don't understand is why the constant rule (that the derivative of a constant is equal to 0) doesn't apply to the 4 and pi in this case.

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3. Originally Posted by 2021econ2012
My calc class is studying basic derivatives, and I'm stuck on a problem.
The problem is f(r) = (4/3) * pi * r3 The answer is 4pir2

$\dfrac{d}{dr}\, \left[\dfrac{4\pi r^3}{3}\right]$

. . .$=\, \dfrac{4\pi}{3}\, \cdot\, \dfrac{d}{dr}\left[r^3\right]$

. . .$=\, \dfrac{4\pi\, \cdot\, 3r^2}{3}$

. . .$=\, 4\pi r^2$

I understand that the power rule calls for the exponent of 3 to be brought to the front and then the exponent is decreased by one. I also understand that the 3 in the denominator cancels out with the 3 in the numerator.
What I don't understand is why the constant rule (that the derivative of a constant is equal to 0) doesn't apply to the 4 and pi in this case.
You're correct about the constant rule, and it does apply in this case, but probably not in the way you're expecting. The addition/subtraction rule for derivatives says that:

$\dfrac{d}{dx} \left( f(x) \pm g(x) \right) = \dfrac{d}{dx} \left( f(x) \right) \pm \dfrac{d}{dx} \left( g(x) \right)$

Logically, it then makes sense that the same would apply for multiplication and you'd have:

$\dfrac{d}{dx} \left( f(x) \cdot g(x) \right) = \dfrac{d}{dx} \left( f(x) \right) \cdot \dfrac{d}{dx} \left( g(x) \right)$

But this is not true. Instead, we have another rule which you might not have learned yet, called the product rule. Using the notation of $f'(x) = \dfrac{d}{dx} \left( f(x) \right)$ for shorthand, it says:

$\dfrac{d}{dx} \left( f(x) \cdot g(x) \right) = f'(x) \cdot g(x) + f(x) \cdot g'(x)$

Letting $f(x) = \dfrac{4\pi}{3} \text{ and } g(x) = r^3$, we can apply the product rule in this case to see what's really going on:

$\dfrac{d}{dx} \left( \dfrac{4\pi}{3} \cdot r^3 \right) = 0 \cdot r^3 + \dfrac{4\pi}{3} \cdot 3r^2 = 4 \pi r^2$

This implementation of the product rule then leads directly to another rule of derivatives, the constant multiple rule, whereby any constant multiple can be "pulled out front" of the derivative.

4. Originally Posted by 2021econ2012
What I don't understand is why the constant rule (that the derivative of a constant is equal to 0) doesn't apply to the 4 and pi in this case.
It's because you're not taking the derivative of 4*Pi.

Look at the second line of your work. You factored out 4*Pi/3. The derivative operator d/dr is applied only to the function r^3.

There's a name for this.

The Constant Multiple Rule: The derivative of a constant times a function is just the constant times the derivative of the function.