Derivatives of constants

The derivative of a constant term is 0, in your case the term isn't constant. The "3" you're probably referring to is a factor which belongs to the term "x²/3".

Since the derivative is lineair, you can put the factor up front and get:

[tex]\left( {\frac{{x^2 }}
{3}} \right)^\prime = \frac{1}
{3}\left( {x^2 } \right)^\prime = \frac{{2x}}
{3}[/tex]

 

Try applying the division rule to this. (of course, it's better if you remember the scalar multiplication rule, but since this is a sticky point for you, try doing it with the division rule and see if you're happy)

 

I don't think that is his problem, Hurkyl!

Because (f+ g)'= f'+ g', (f+ Constant)'= f'+ Constant'= f'+ 0= f' so you can "ignore" the constant. (Notice the quotes- you are not IGNORING it! You are THINKING about it, recognizing that its derivative is a constant and so "adding 0".

Because (fg)'= f'g+ fg', (Constant*g)'= (Constant)'g+ (Constant)g'= 0*g+ Constant*g'. The derivative of a constant times a function is the constant times the derivative of the function. I guess you could say you are "ignoring" the constant here but that's why I don't like that word! THINK about what you are doing!

Surely you know that every function involves some constants (even if they are 1). You can't just "ignore" them.

The two rules I mentioned above: (f+ Constant)'= f' and (Constant*f)'= Constant*f' are as close as you can come to "ignoring" the Constant!

 

I'm hoping he'll appreciate what's going on better if he does the division rule and sees the constant resurface at the end. In effect, I'm hoping he'll work out for himself exactly what you told him about the product rule.

 

i'm not seeing a constant there. a quick mental calculation gives [tex]\frac{d}{dx}f(x)=\frac{2}{3}x^{-\frac{1}{3}}[/tex]

 

that's why you use [tex]\LaTeX[/tex]

 

And, of course, the strict interpretation x^2/3 is (x^2)/3.

 

then it's [tex]\frac{d}{dx}f(x)=\frac{2}{3}x[/tex]

 

I was thinking that hotrocks007 meant x^(2/3) but was thinking "Aha- there's a constant, 2/3, in there and the derivative of a constant is 0! So the derivative of this is 0!" That was the reason for my remark that every function contains "constants"!