# Differentiating x^y + y^x + (lnx)^x etc

1. Apr 26, 2004

### pig

How can this be done?

I don't even know how I would begin.. How would you differentiate stuff like x^(y^(x^y))? Where y is a function of x, not a constant of course..

2. Apr 26, 2004

### arildno

The example:
x^(y^(x^y))=x^(z(x))

d/dx x^(z(x))=z(x)x^(z(x)-1)+x^(z(x))*ln(x)*z'(x)..

3. Apr 26, 2004

### pig

I don't understand this :(

d/dx x^(z(x))=z(x)x^(z(x)-1)+...

Where does the + come from?

I know I should use the chain rule somehow but I can't seem to figure out how.. I'm having problems with differentiating both x^(f(x)) and f(x)^x..

4. Apr 26, 2004

### pig

Hmm if I do this:

f(x) = x^y
ln f(x) = ln x^y
ln f(x) = ylnx
(ln f(x))' = (ylnx)'
f'(x)/f(x) = y'lnx + y/x
f'(x) = f(x)*(y'lnx + y/x)

(x^y)' = x^y*(y'lnx + y/x)

Is this right?

5. Apr 26, 2004

### pig

(x^y)' = x^y*(y'lnx + y/x)

(x^y)' = yx^(y-1)+x^y*lnx*y'

I think I understand what you wrote after all.. Thanks arildno :)

Last edited: Apr 26, 2004
6. Apr 26, 2004

### NateTG

Use the chain rule:

$$\frac{d}{dx} f(g(x))=g'(x)f'(g(x))$$

So
$$\frac{d}{dx} x^{(y^{(x^y)})}= \frac{d}{dx} e^{(\ln x \times y^{(x^y)})} = \frac{d}{dx} (\ln x \times y^{(x^y)}) \times x^{(y^{(x^y)})}$$
$$=(\frac{y^{(x^y)}}{x} + \ln x \times \frac{d}{dx} y^{(x^y)}) \times x^{(y^{(x^y)})}$$
$$=(\frac{y^{(x^y)}}{x} + \ln x \times \frac{d}{dx} e^{(y \times x^y)}) \times x^{(y^{(x^y)})}$$
$$=(\frac{y^{(x^y)}}{x} + \ln x \times \frac{d}{dx} {(y \times x^y)} \times y^{(x^y)}) \times x^{(y^{(x^y)})}$$
$$=(\frac{y^{(x^y)}}{x} + \ln x \times (y' x^y + y \frac{d}{dx} (x^y)) \times y^{(x^y)}) \times x^{(y^{(x^y)})}$$
$$=(\frac{y^{(x^y)}}{x} + \ln x \times (y' x^y + y \frac{d}{dx} (y \ln x ) \times x^y) \times y^{(x^y)}) \times x^{(y^{(x^y)})}$$

$$=(\frac{y^{(x^y)}}{x} + \ln x \times (y' x^y + y (\frac{y}{x} + xy' ) \times x^y) \times y^{(x^y)}) \times x^{(y^{(x^y)})}$$
Obviously, some regrouping is necessary. I somehow doubt that that's the same as what arnildo had.

7. Apr 27, 2004

### arildno

Well, I didn't bother to compute z'(x).
Here's the simplest way to compute x^(z(x)):

F(x,z)=x^(z), H(x)=F(x,z(x)).

dH/dx=d^(p)F/dx+d^(p)F/dz*z'(x), where d^(p)/dx is the partial derivative with respect to x.