# Derivative question

1. May 29, 2006

### Mr. Snookums

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

I'm not sure how to do this. Any tips? How would I treat the power?

Should be x^3 as the power, not 3x.

2. May 29, 2006

### danago

For that, its probably easiest to simply multiply out the bracket, which would give you an expression with two terms, then differentiate those two terms separately.

Last edited: May 29, 2006
3. May 29, 2006

### neutrino

His tex code seems to suggest that x^2+1 is raised to x^3. In this case, you can easily solve it by taking the logarithm and then differentiate (I find it simpler that way).

4. May 29, 2006

### danago

oh. my mistake then

5. May 29, 2006

### VietDao29

Okay, I assume that you mean:
$$y = (x ^ 2 + 1) ^ {x ^ 3}$$
Ok, I'll give you an example:
Example:
Differentiate xx with respect to x.
There are 2 ways to solve it.
------------
The first way is to try to change xx into esomething. Then from there, we can just use the Chain Rule to arrive at what the problem asks.
$$x ^ x = \left( e ^ {ln x} \right) ^ x = e ^ {x \ln x}$$
Now we have:
$$\left( x ^ x \right)' = \left( e ^ {x \ln x} \right)' = (x \ln x)' e ^ {x \ln x} = (\ln x + 1) e ^ {x \ln x} = x ^ x (\ln x + 1)$$
------------
The second way is to let y = xx
Now take the natural log of both sides, we have: ln y = ln(xx) = x ln x.
Now differentiate both sides with respect to x gives:
$$\frac{y'_x}{y} = \ln x + 1$$
$$\Rightarrow y'_x = y( \ln x + 1)$$
$$\Rightarrow y'_x = x ^ x ( \ln x + 1)$$
------------
The two methods give the same result. Now can you go from here? :)

6. May 29, 2006

### dav2008

Just to expand a bit on the second way and why you can do that...

In your original problem you are looking for $$\frac{d}{dx}((x ^ 2 + 1) ^ {x ^ 3})$$

Note that there is no y involved anywhere in the expression, but if you introduce a "dummy variable" y and set it equal to the given expression you get $$y=(x ^ 2 + 1) ^ {x ^ 3}$$

Notice now that $$\frac{dy}{dx}=\frac{d}{dx}((x ^ 2 + 1) ^ {x ^ 3})$$ In other words, if you can manipulate the new equation ($y=(x ^ 2 + 1) ^ {x ^ 3}$)and solve for $\frac{dy}{dx}$ then you will have solved for the initial derivative.

Last edited: May 29, 2006
7. May 29, 2006

### Mr. Snookums

Ah, yes. I had forgotten all about solving these buggers with logarithms. Thank you.

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