# Homework Help: Need help with a derivative

1. Jul 14, 2012

### mesa

Lets say we have this:
y=(x^x)^x
what is y'=?

I can handle the variable to a variable but am not sure how to approach it when another variable is nested inside the function

2. Jul 14, 2012

### Saitama

You can write the expression as $y=x^{x^{2}}$, take logarithm on both the sides and find the derivative.

3. Jul 14, 2012

### mesa

Sorry I was trying to put it in more generic form, lets try it this way:
y=(3x^2x+1)^x+1

4. Jul 14, 2012

### Saitama

You can still do the same, take logarithm on both sides.

5. Jul 14, 2012

### mesa

Thats where I am getting stuck, how do I take the ln of both sides with the inner function also being a variable to a variable?

6. Jul 14, 2012

### Saitama

When you take logarithm of both the sides, you get an equation like this:
$$\ln y=(x+1)\ln (3x^{2x}+1)$$
Differentiate it using the product rule, when its turn to differentiate 3x^(2x), calculate its derivative separately and use that.

7. Jul 14, 2012

### mesa

Which gets me to here:

(1/y)(dy/dx)=(x+1)[1/(3x^2+1)]...

and now we where I am stuck lol, how do I differentiate (3x^2+1) by itself?

8. Jul 14, 2012

### Saitama

What's the derivative of $y=3x^{2x}$?

9. Jul 14, 2012

### mesa

Hah ha, yes I did :)
The derivative of y=3x^2x?
Don't know

10. Jul 14, 2012

### Saitama

You know it, don't you? Take logarithm on both sides again.

11. Jul 14, 2012

### mesa

So we can just set up a new function for the 3x^2x and set equal to y then differentiate?

So it would be:

(1/y)(dy/dx)=(2)+(ln 3x)(2)

12. Jul 14, 2012

### Saitama

No.
Let's first calculate the derivative z=3x^(2x) (i use z because we have already used y).
Differentiating y w.r.t. x, we get:
$$\frac{1}{y}\frac{dy}{dx}=\ln (3x^{2x}+1)+(x+1)\frac{1}{3x^{2x}+1}\frac{d}{dx}(3x^{2x}+1)$$

Calculate derivative of 3x^(2x) or derivative of z separately.

13. Jul 14, 2012

### mesa

Mind if we get rid of that +1 from the original eqation? It was meant to be part of the exponent but I miswrote it

So we get:
z=3x^(2x)

lnz=(2x)ln(3x)
differntiate to:
(1/z)(dz/dx)=(2x)(1/(3x)(3)+(2)ln(3x)

14. Jul 14, 2012

### Reptillian

Use the Chain Rule and remember that d/dx(ax) = axln(a) where a is a constant

15. Jul 14, 2012

### mesa

Yeah, I was able to get that far but how do we differentiate when there is a variable to a variable power inside a variable to a variable power function?

16. Jul 14, 2012

### Ray Vickson

I don't know what your expression means. Is it
$(3 x^2 x + 1)^x + 1, \; (3 x^{2x}+1)^x + 1,\; (3 x^{2x} + 1)^{x+1},$ or something else? If I read it *using standard rules* it means the first. If you mean the last, use brackets: (3 x^(2x) + 1)^(x+1).

RGV

17. Jul 14, 2012

### mesa

yeah, I knida screwed the pooch on that one, it was supposed to be:

(3x^(2x+1))^(x+1)

18. Jul 14, 2012

### Saitama

If it's this, then it should be easy to solve.
You can write this as:
$y=3x^{2x^2+3x+1}$
Take logarithm on both sides and find the derivative.

19. Jul 14, 2012

### mesa

Yup, but I need a solution without simplification

20. Jul 14, 2012

### SammyS

Staff Emeritus
Keep applying derivative rules including the chain rule, from the "outside" towards the "inside".

For instance:

$\displaystyle \frac{d}{dx}\left(f\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\right)$
$\displaystyle = f'\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot \frac{d}{dx}\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)$

$\displaystyle = f'\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot g'\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot \frac{d}{dx}\left(h\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\right)$

$\displaystyle = f'\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot g'\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot h'\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\cdot \frac{d}{dx}\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)$

$\displaystyle = f'\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot g'\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot h'\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\cdot \left( s'\left(u(x)\right)\cdot\frac{d}{dx}u(x)+ \frac{d}{dx}\left(v(x) \cdot w(x)\right)\right)$

$\displaystyle = f'\left(g\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\right)\cdot g'\left(h\left(s\left(u(x)\right)+v(x)\cdot w(x)\right)\right)\cdot h'\left(s\left(u(x)\right)+v(x) \cdot w(x)\right)\cdot \left(s'\left(u(x)\right)\cdot u'(x)+ \left(v'(x) \cdot w(x)+v(x) \cdot w'(x)\right)\right)$