# Differentiation problem!

• Mar 30th 2010, 02:04 AM
watto23
Differentiation problem!
Hiya people, I'm a bit stuck on a problem and I have to differentiate this function:

f(x) = 2^(x^2)

i get f '(x) = [2x^3][2^((x^2)-1)] using the chain rule

however one of my mates has got a different answer (and he's a bit more reliable than me!):
f '(x) = [2x][2^(x^2)][ln(2)]

Could anyone please help me on this and possibly explain the derivation as I have to write down the steps in the process
thanks, Watto
• Mar 30th 2010, 02:18 AM
drumist
Consider

$\displaystyle (2x)^4$

In this case, the x term(s) appear in the base, and the exponent is a constant. For this, you use the rule where you subtract one from the exponent and multiply the whole thing by the derivative of the base.

Now consider

$\displaystyle (4)^{2x}$

In this case, the x term(s) appear in the exponent, and the base is constant. This uses an entirely separate rule. For this, you should copy the expression down (without subtracting 1 from the exponent), and multiply it by the derivative of the exponent, and multiply that by the natural log of the base.

Your friend's answer is correct. Can you figure out how he got it now?
• Mar 30th 2010, 02:23 AM
sa-ri-ga-ma
If f(x) = a^g(x) , then f'(x) = g'(x)*a^g(x)*lna
• Mar 30th 2010, 02:28 AM
watto23
thanks i get it now! I appreciate your help
• Mar 30th 2010, 02:44 AM
mr fantastic
Quote:

Originally Posted by watto23
thanks i get it now! I appreciate your help

You probably do. Nevertheless, for those in the cheap seats, note that $\displaystyle 2^{x^2} = e^{\ln 2^{x^2}} = e^{x^2 (\ln 2)} = e^{kx^2}$ where $\displaystyle k = \ln 2$ and the chain rule can now be easily applied.