Derivative of a constant

[MarkSA]

This is an odd question, but here goes.

Say I have:
f(x) = 5^3

The derivative of this is 0, because the derivative of a constant is zero... However... Let's say that I was actually working this problem out for some reason...

Would it be technically incorrect (calculus-wise) to apply the power rule to this first and then the chain rule? (I know it would be pointless, when a person can just use the constant rule)

For instance, say I worked out f(x) = 5^3 like so using the power rule first and then the chain rule:
f'(x) = [3(5)^2] * [0] = 0
With the [0] meaning 0 is the derivative of 5.

Is this, by the calculus rules, a valid way of doing this problem, albeit long, or is it incorrect? (Can you explain why?)

 

[galactus]

Here's a better way to look at it, perhaps:

Let

\(\displaystyle \L\\f'(x)=\lim_{h\to\0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to\0}\frac{c-c}{h}=\lim_{h\to\0}[0]=0\)

If f(x)=c is a constant function, then the graph of f is a line parallel to the x-axis and, therefore, the tangent line at each point is horizontal.

So, \(\displaystyle \L\\f'(x)=m_{tan}=0\), since a horizontal line has slope 0.

It seems rather redundant. f(x)=125?. 125 is not a function in x. It is a constant which has a derivative of 0. A derivative is a slope at a point.
If you had \(\displaystyle 5x^{3}\), then you can find the slope at any point by finding the derivative, \(\displaystyle 15x^{2}\). 5^3 has no slope.
By your method, you have \(\displaystyle f'(x)=3(5)^{2}\)
This would mean \(\displaystyle f(x)=x^{3}\) and you found the derivative at x=5. Multiplying by 0 is not relevant.

 

[MarkSA]

Hm, i'm still a little confused.. Let me explain where my thinking on this is coming from..

If I was to use some of the derivative rules on a function, say:

f(x) = 5x^3

The derivative is 15x^2
If I work it out this way:
5 * [3(x)^2] * [derivative of x which equals 1] = 15x^2
Isn't it correct by the derivative rules, though redundant, to include * 1 in place of the derivative of x, which indicates the derivative of what's inside the parentheses when using the chain rule? I'd always assumed this step, multiplying by * 1, was just omitted because showing multiplication by 1 is rather pointless.

So, if what I said above is correct, then can't this be applied to the derivative of any function where the chain rule is used? Like in my first example of f(x) = 5^3. Doing the same thing, except 5 is inside the parentheses instead of x:
3(5)^2 * [derivative of 5 which equals 0] = 0

I'm getting a similar impression from doing derivatives of trig functions like:
f(x) = sin(5x)
Derivative: cos(5x) * [derivative of 5x equals 5] = 5cos(5x)
vs
f(x) = sin(x)
Derivative: cos(x) * [derivative of x equals 1] = cos(x)
vs
f(x) = sin(5)
Derivative: cos(5) * [derivative of 5 equals 0] = 0

Again, doing it this way is very redundant (and rather pointless too I know), but what i'm wondering is if this is legally (by the calculus rules), -a- correct way of evaluating the derivatives of these functions? Can I literally say that I did these derivatives correctly, even if I took the ridiculously long path? If not, could you explain where i'm going wrong with them?

 

[skeeter]

I'll agree with pointless and redundant, but technically, legal and correct.

using the chain rule for \(\displaystyle \L y = [f(x)]^n\) ...

\(\displaystyle \L y' = n[f(x)]^{n-1} \cdot f'(x)\)

if f(x) is a constant function, then the derivative factor f'(x) is indeed 0, making y' = 0.

 

[MarkSA]

Thank you! That just about clears it up then...

These are probably strange questions, but there's a method behind the madness. Using the chain rule that way is how I taught myself on my own. However, my class was taught differently, and I recently had an answer counted wrong on the test by using this method.

The problem was similar to:
Find the derivative of:
f(x) = pi^3

The answer is 0 of course, but i'm paranoid about showing work, so I used the same method as in the above post using the power and chain rules to solve it by showing:
f'(x) = 3*(pi)^2 * [derivative of pi is 0] = 0

The grader said: "The derivative of a constant is 0, it's not 3 pi squared multiplied by 0"

So i'm trying to understand if i'm using the correct method here or if i'm missing something. The only thing is that in the book, it says the chain rule deals with a composite of functions. Can the pi in 3*(pi)^2 be considered a function? I wouldn't think so, but... using the chain rule on what's in the parentheses, I still end up with the same end answer. So it's still technically correct (and won't come back to haunt me down the road) to use the chain rule in this fashion on whatever is inside the parentheses, including constants?

 

[tkhunny]

It seems to me we are missing the point, just a bit. What is the "Power Rule". What it is depends on how it was defined.

If the working definition was this:

For any real number, a, if \(\displaystyle f(x)\;=\;x^{a},\;then\;f'(x)\;=\;a*x^{a-1}\)

Then the application to \(\displaystyle f(x)\;=\;\pi^{n}\) simply is wrong. This structure does not fit the definition. The rule is not applicable. Any attempt to use the rule should be given less than total credit.

 

[MarkSA]

Yes, that's how it's defined in the book.

Perhaps they just never thought to adjust the power rule to encompass constants too? (Ok, i'm stretching here I know)

I see why it was counted wrong now (although it does sting having the correct 'answer' but worked out incorrectly.)

So, I guess it's just an unfortunate coincidence that applying the power rule to a function with just a constant equals the right answer (0)?