- #1
autodidude
- 333
- 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?
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?