Which method is correct for finding the derivative of y = x^x - C and why?

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In summary, there are two methods for finding the derivative of y = x^x - C. Method 1 involves taking the logarithm of both sides and using the Product Rule, while Method 2 involves using implicit differentiation. Both methods produce different answers, and it is unclear which one is correct. However, after some discussion and clarification, it is determined that Method 2 is the correct method.
  • #1
Hertz
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I promise guys, no homework here, just curiosity.

I am trying to find dy/dx for the equation y = x^x - C, where C is any arbitrary constant. I've found two ways that SHOULD be ok to take this derivative, but they produce different answers, I was wondering which method is correct and which method is incorrect. Also, why? It seems to me that both of these methods should be ok. Anyways, here they are:

Method 1:
[itex]y = x^x - C[/itex]
[itex]ln(y) = ln(x^x - C)[/itex]
[itex]ln(y) = \frac{x ln(x)}{ln(C)}[/itex]
Now take the derivative:
[itex]\frac{\frac{dy}{dx}}{y}=\frac{1}{ln(C)}(x ln(x))'[/itex]
Using the Product Rule, it can be seen that [itex](x ln(x))' = ln(x) + 1[/itex]. Therefore:
[itex]\frac{dy}{dx}=\frac{y}{ln(C)}(ln(x) + 1)[/itex]
[itex]\frac{dy}{dx}=\frac{x^x}{ln(C)}(ln(x) + 1)[/itex]

Method 2:
[itex]y = x^x - C[/itex]
[itex]y' = (x^x)' - C'[/itex]
[itex]y' = (x^x)'[/itex]
[itex](x^x)'[/itex] can be evaluated using method 1 for the equation [itex]y = x^x[/itex]
[itex]\frac{dy}{dx} = x^x(ln(x) + 1)[/itex]

Method one seems a bit less hand wavy, so I'm more confident in it; however, the derivative shouldn't depend on C, so that makes me lean more toward Method 2.

Anybody have any input they'd be willing to share?
 
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  • #2
rewrite x as e^ln(x) to e^(xln(x)) and then use the rule d/dx (e^u) = e^u * u'
 
  • #3
[itex]\ln(a-b)\neq\ln a/\ln b[/itex]. If I were you I'd just try to rewrite the [itex]x^x[/itex] as [itex]e^{f(x)}[/itex].
 
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  • #4
jedishrfu said:
rewrite x as e^ln(x) to e^(xln(x)) and then use the rule d/dx (e^u) = e^u * u'

Rewrite which x? Both? In the original or in the derivative?

Are you saying to do this?:

[itex]y = x^x - C[/itex]
[itex]e^y = e^{x^x - C} = e^{x^x}/e^C[/itex]

And then differentiate?

Ibix said:
[itex]\ln(a-b)\neq\ln a/\ln b[/itex]. If I were you I'd just try to rewrite the [itex]x^x[/itex] as [itex]e^{f(x)}[/itex].

Oh right! I was getting the rule backwards it seems.

So would method 2 be the one that is correct then?

y = x^x can be differentiated pretty easily with implicit differentiation.
 
  • #5
jedishrfu said:
rewrite x as e^ln(x) to e^(xln(x)) and then use the rule d/dx (e^u) = e^u * u'

Hertz said:
Rewrite which x? Both? In the original or in the derivative?

Are you saying to do this?:

[itex]y = x^x - C[/itex]
[itex]e^y = e^{x^x - C} = e^{x^x}/e^C[/itex]

And then differentiate?

jedishrfu is saying that
[itex]x = e^{\ln x}[/itex]
so
[itex]x^x = \left(e^{\ln x}\right)^x = e^{x\ln x}[/itex]

So what you need to take the derivative of is this:
[itex]y = e^{x\ln x} - C[/itex]
and you won't need to worry about implicit differentiation.
 
  • #6
eumyang said:
jedishrfu is saying that
[itex]x = e^{\ln x}[/itex]
so
[itex]x^x = \left(e^{\ln x}\right)^x = e^{x\ln x}[/itex]

So what you need to take the derivative of is this:
[itex]y = e^{x\ln x} - C[/itex]
and you won't need to worry about implicit differentiation.

Oh I see, well let's try:

[itex]y = e^{x\ln x} - C[/itex]
[itex]\frac{dy}{dx} = (e^{x\ln x})'[/itex]
[itex]\frac{dy}{dx} = e^{x\ln x}(ln(x) + 1)[/itex]
[itex]\frac{dy}{dx} = x^x(ln(x) + 1)[/itex]

Thanks for the help guys. I think the only problem I was having was mixing up my logarithm properties :)
 
  • #7
The derivative of an added or subtracted constant is 0 so the derivative of
[itex]y= x^x- C[/itex] is exactly the same as the derivative of [itex]y= x^x[/itex]. Now, y= x ln(x) so dy/dx= ln(x)- (x/x)= ln(x)- 1.
 
  • #8
HallsofIvy said:
The derivative of an added or subtracted constant is 0 so the derivative of
[itex]y= x^x- C[/itex] is exactly the same as the derivative of [itex]y= x^x[/itex]. Now, y= x ln(x) so dy/dx= ln(x)- (x/x)= ln(x)- 1.
I think you meant: ln(y)= x ln(x) so (dy/dx)/y = ln(x) + (x/x)= ln(x)+ 1 (Note: +, not -).
This leads to the same answer as in the OP by Method 2.
Method 1 went wrong after the second line. ln(A-B) is not ln(A)/ln(B). Must be thinking of ln(A/B) = ln(A) - ln(B) (or, equivalently, exp(A-B) = exp(A)/exp(B)).
 

FAQ: Which method is correct for finding the derivative of y = x^x - C and why?

What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It measures how much a function changes in response to a small change in its input variable.

Why is it important to take derivatives?

Derivatives are essential in many areas of science and engineering, including physics, economics, and statistics. They allow us to analyze the behavior of functions, solve optimization problems, and make predictions about the future.

How do I take a derivative?

To take a derivative, you can use the basic rules of differentiation, such as the power rule, product rule, quotient rule, and chain rule. These rules allow you to find the derivative of any function by breaking it down into simpler parts and applying the appropriate rule.

What is the difference between a derivative and an integral?

A derivative measures the rate of change of a function, while an integral measures the accumulation of a function over an interval. In other words, a derivative tells you how fast something is changing, and an integral tells you how much of something there is.

Can I use a calculator to take derivatives?

Yes, many scientific and graphing calculators have built-in functions for finding derivatives. However, it is important to understand the underlying concepts and rules of differentiation to use these tools effectively.

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