# Help with derivatives of these functions

• Nov 26th 2006, 11:32 AM
Dr. Noobles
Help with derivatives of these functions
1. f(x)= e^(2*pi*x)
2. f(x)= x^(2*pi*e)
3. f(x)= (e*pi)^(2x)
4. f(x)= pi^(e)^(2x)

thanks
• Nov 26th 2006, 11:37 AM
topsquark
Quote:

Originally Posted by Dr. Noobles
1. f(x)= e^(2*pi*x)
2. f(x)= x^(2*pi*e)

Recall that $\displaystyle \frac{d}{dx}e^{x} = e^{x}$.

The first is going to be done by the chain rule:
1. $\displaystyle f(x)= e^{2 \pi x}$

$\displaystyle f'(x) = e^{2 \pi x} \cdot 2 \pi = 2 \pi e^{2 \pi x}$

Since $\displaystyle \pi$ and "e" are just constants, the second one is familiar to you already. It is just the power law:
$\displaystyle \frac{d}{dx}x^n = nx^{n-1}$

2. $\displaystyle f(x)= x^{2 \pi e}$

$\displaystyle f'(x) = (2 \pi e) x^{2 \pi e - 1}$

-Dan
• Nov 26th 2006, 11:41 AM
topsquark
Quote:

Originally Posted by Dr. Noobles
3. f(x)= (e*pi)^(2x)

Since $\displaystyle \pi$ and "e" are constants this is the derivative of a constant to a function:
$\displaystyle \frac{d}{dx}a^x = ln(a) \cdot a^x$

(We also need to use the chain rule.)

$\displaystyle f(x) = (e \pi)^{2x}$

$\displaystyle f'(x) = \left ( ln(e \pi) \cdot (e \pi)^{2x} \right ) \cdot 2$

$\displaystyle f'(x) = 2 \left ( ln(e) + ln( \pi ) \right ) (e \pi )^{2x}$ Using a property of logarithms.

$\displaystyle f'(x) = 2 \left ( 1 + ln( \pi ) \right ) (e \pi )^{2x}$

-Dan
• Nov 26th 2006, 11:48 AM
topsquark
Quote:

Originally Posted by Dr. Noobles
4. f(x)= pi^(e)^(2x)

We need some clarification here. Is your function:
$\displaystyle f(x) = \pi ^{ \left ( e^{2x} \right ) }$

or

$\displaystyle f(x) = \left ( \pi ^{e} \right )^{2x}$

For the first case, we need to use the chain rule a few times:
$\displaystyle f(x) = \pi ^{ \left ( e^{2x} \right ) }$

$\displaystyle f'(x) = \left ( ln( \pi ) \cdot \pi ^{ \left ( e^{2x} \right ) } \right ) \cdot \left ( e^{2x} \right ) \cdot (2)$

$\displaystyle f'(x) = 2 ln( \pi ) e^{2x} \cdot \pi ^{ \left ( e^{2x} \right ) }$

In the second case note that $\displaystyle \left ( \pi ^{e} \right )^{2x} = \pi ^{2ex}$, so
$\displaystyle f(x) = \left ( \pi ^{e} \right )^{2x} = \pi ^{2ex}$

$\displaystyle f'(x) = \left ( ln( \pi ) \cdot \pi ^{2ex} \right ) \cdot (2e)$

$\displaystyle f'(x) = 2e \cdot ln( \pi ) \cdot \pi ^{2ex}$

-Dan