# Differentiating y=x^x

1. Aug 15, 2012

### autodidude

Differentiating y=x^x

x=ln(y)

I changed the base to e

$$x=\frac{ln(y)}{ln(x)}$$
$$xln(x) = ln(y)$$
$$e^{xln(x)} = y$$
$$e^{xln(x)}(1+ln(x) = \frac{dy}{dx}$$

The answer the calculator got was $$x^{x(1+ln(x))}$$ so I noticed that since $$y=x^x$$ and $$e^{xln(x)} = y$$, 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?

2. Aug 15, 2012

### Dickfore

yes, but the factor $1+\ln x$ should not be in the exponent (chain rule).

3. Aug 15, 2012

### autodidude

^ Thank you...yeah I made an error there