Can Anyone Help Me Solve Derivative x^x^x with Additional X?

  • Thread starter silence
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In summary, the conversation is discussing the topic of solving x^(x^x) and the use of the product rule and chain rule to differentiate it. The conversation also touches on the possibility of solving (x^x)^x and the importance of not re-opening old discussions.
  • #1
silence
i can solve x^x but adding this new x just confuses me any help will do, X^x^x
 
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  • #2
Do you mean (x^x)^x or x^(x^x)? I'm assuming the former.
 
  • #3
sorry i meant x^(x^x)
 
  • #4
Glad you picked that one. Get something like ln(y) = xxln(x) and use the product rule. Don't forget you already know d(xx)/dx :wink:
 
  • #5
wow i never noticed that i was doing it a long way which would have come out wrong anyways. thanks for the help
 
  • #6
ok ok
 
Last edited:
  • #7
djuiceholder said:
but can you do (x^x)^x ??
it seems impossible

1. It is not at all imposible; note that this equals: [tex](x^{x})^{x}=x^{x^{2}}[/tex]
Rewriting this as:
[tex]x^{x^{2}}=e^{x^{2}\ln(x)}[/tex]
We may readily differentate this by means of the chain rule, yielding the derivative:
[tex]x^{x^{2}}(2x\ln(x)+x)[/tex]

2. Please do not re-open nearly 6-year old threads.
 
  • #8
ok ok
 
Last edited:

1. What is the derivative of x^x^x?

The derivative of x^x^x is a complex function that involves multiple applications of the chain rule and logarithmic differentiation. The result is x^x^x * (1 + ln(x) + ln(ln(x))).

2. How do you find the derivative of x^x^x?

To find the derivative of x^x^x, you need to use logarithmic differentiation and the chain rule. First, take the natural logarithm of both sides to get ln(y) = x^x * ln(x). Then, use the product rule and chain rule to simplify and solve for y'. Finally, replace y with x^x^x to get the final derivative.

3. What is the graph of the derivative of x^x^x?

The graph of the derivative of x^x^x is a non-linear function with a steep incline at x=1 and a horizontal asymptote at x=0. It also has a critical point at x=1, where the derivative is undefined.

4. Is the derivative of x^x^x continuous?

No, the derivative of x^x^x is not continuous. It has a discontinuity at x=0, where the derivative is undefined.

5. What is the significance of x=1 in the derivative of x^x^x?

The value of x=1 is a critical point in the derivative of x^x^x, where the derivative is undefined. This means that the original function, x^x^x, has a point of inflection at x=1, where the concavity changes from upward to downward.

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