Differentiating y=x^x: Acceptable Method?

  • Context: Graduate 
  • Thread starter Thread starter autodidude
  • Start date Start date
  • Tags Tags
    Differentiating
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
autodidude
Messages
332
Reaction score
0
Differentiating y=x^x

x=ln(y)

I changed the base to e

[tex]x=\frac{ln(y)}{ln(x)}[/tex]
[tex]xln(x) = ln(y)[/tex]
[tex]e^{xln(x)} = y[/tex]
[tex]e^{xln(x)}(1+ln(x) = \frac{dy}{dx}[/tex]

The answer the calculator got was [tex]x^{x(1+ln(x))}[/tex] so I noticed that since [tex]y=x^x[/tex] and [tex]e^{xln(x)} = y[/tex], then I could replace it with x^x in the final answer

Is this an acceptable method? Is there any circular logic I missed? Could I leave it as is wihtout writing x^x?
 
Physics news on Phys.org
yes, but the factor [itex]1+\ln x[/itex] should not be in the exponent (chain rule).
 
^ Thank you...yeah I made an error there