# Using chain rule to differentiate x^x

1. Oct 7, 2009

### Shakas

1. The problem statement, all variables and given/known data

Use the chain rule to find (d/dx)(xx) by using the function f(y,z)=yz.

2. Relevant equations

Chain rule: $$\frac{dz}{dt}$$ = $$\frac{\partial z}{\partial x}$$ $$\frac{dx}{dt}$$ + $$\frac{\partial z}{\partial y}$$ $$\frac{dy}{dt}$$

3. The attempt at a solution

I honestly have no clue on how to use chain rule in this problem. I have spent about an hour trying to set up a chain rule equation and I have given up hope on finding one.

The partial derivatives of f(y,z) are:
$$\frac{\partial x}{\partial y}$$ = zyz-1
$$\frac{\partial x}{\partial z}$$ = yzln(y)

I have a strong feeling that I need to set up parametric equations somewhere to cancel out and get dy/dx, but I can't seem to find any possible way.

I know how to find the derivative using one-variable calculus, and it comes to out be [dy/dz = xx(1+ln(x))]. Any hints on how to setup a chain rule equation would be greatly appreciated, thank you.

Last edited: Oct 7, 2009
2. Oct 7, 2009

### slider142

Let y(t) = t and let z(t) = t. What you're looking for is then df/dt, which the fits the chain rule you posted nicely.

3. Oct 8, 2009

### Shakas

Alright so taking it a step further:

$$\frac{df}{dt} = \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}$$

$$y(t)=t$$

$$z(t)=t$$

$$\frac{dy}{dt} = 1$$

$$\frac{dz}{dt} = 1$$

$$\frac{df}{dt} = \frac{\partial f}{\partial y}$$ + $$\frac{\partial f}{\partial z}$$

$$\frac{df}{dt} = zy^z-1 + y^z ln(y)$$

$$\frac{df}{dt} = \frac{zy^z}{y} + y^z ln(y)$$

$$\frac{df}{dt} = y^z (\frac{z}{y} + ln(y))$$

$$y=x$$

$$z=x$$

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

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

Thank you very much for your help.

4. Oct 8, 2009

### blake knight

The answer to your problem: (d/dx)(x^x) = y(x^x)....z is to be cancelled out along the operations..try differentiating both sides of the equations.

First let y=x^x and
f(y,z)=u=y^z--->from here solve for y in terms of z, you can do that by getting the ln of both sides.
You will arrive at having: y=e^lnu.

Hope this could be of any help.