# derivative of y=x^x

#### spezialize

##### New member
can someone explain how to get the deriv of y=x^x.. I think you have to use log rules but I am unsure how to go about this.

#### Unco

##### Senior Member
$$\displaystyle y = x^x$$

Take natural logs

$$\displaystyle \ln{y} = \ln{\left(x^x\right)}$$

Log rule: $$\displaystyle \ln{\left(x^n\right)} = n\ln{x}$$:

$$\displaystyle \ln{y} = x\ln{x}$$

Differentiate implicitly wrt x, apply the product rule to the RHS:

$$\displaystyle \frac{1}{y} \frac{dy}{dx} = \ln{x} + 1$$

Remember that we had $$\displaystyle y = x^x$$ so we get

$$\displaystyle \frac{dy}{dx} = x^x(\ln{x} + 1)$$

#### pka

##### Elite Member
$$\displaystyle y = x^x$$ leads to $$\displaystyle \ln (y) = (x)\ln (x)$$.

Differentiate implicitly: $$\displaystyle \frac{{y'}}{y} = \ln (x) + 1$$.

Solve: $$\displaystyle y' = y(\ln (x) + 1) = x^x (\ln (x) + 1)$$

#### spezialize

##### New member
thank you very much for explaining that. Makes sense.