Another problem having to do with related rates

Question

lifeisadangerousgame · Answer

Is 4/3 a constant too, why it doesn't go to 0 when you take the derivative?

3mar · Answer

Because it does not exist alone, it is a coefficient of the variable being derived!

solomonzelman · Answer

It is a coefficient of a function.

solomonzelman · Answer

$\color{black}{\displaystyle \frac{d}{dt}(3\cdot f(t))=3\cdot f'(t)  }$      Right?

solomonzelman · Answer

We say that the derivative of a constant is 0, when a constant (alone) is differentiated. (Not when a constant is a coefficient of some function, but in a case like below,
(example)    $\color{black}{\displaystyle \frac{d}{dt}(\pi)=0  }$.

solomonzelman · Answer

$\color{black}{\displaystyle \frac{d}{dt}\left[\frac{4}{3}\pi t^3\right]=3\cdot \frac{4}{3}\pi t^{3-1}=4\pi t^{2}   }$       Right? (By power rule)

solomonzelman · Answer

$\color{black}{\displaystyle \frac{d}{dt}\left[\frac{4}{3}\pi [{\tt r}(t)]^3\right]=3\cdot \frac{4}{3}\pi [{\tt r}(t)]^{3-1}\cdot {\tt r}'(t)=4\pi t^{2}{\tt r}'(t)   }$

by power rule, and chain rule

solomonzelman · Answer

Questions ?

lifeisadangerousgame · Answer

I still don't understand how it's a coefficient of the function. Isn't it just like having 3x^2?

solomonzelman · Answer

$\color{black}{\displaystyle \frac{d}{dx}(3x^2)=3x^{2-1}\times 2= 6x }$

Notice that this $3$ doesn't just go to 0. Rather, it just contributes to the derivative of $x^2$, because you can as well note that,

$\color{black}{\displaystyle \frac{d}{dx}(3x^2)=3\left[\frac{d}{dx}(x^2)\right]=3\left[2x\right]=6x }$

solomonzelman · Answer

So, yes, it is a coefficient, just like in case $\color{red}{3}x^2$ that you mentioned.

lifeisadangerousgame · Answer

Okay, I just needed to see it in relation of something I already understood. Thank you so much, it makes sense!

solomonzelman · Answer

Oh, ok:)
 yw!