# Simple Differentiation

• Aug 27th 2009, 05:38 PM
xwrathbringerx
Simple Differentiation
How do I exactly differentiate 3^(5x)

Thanxx a lot!
• Aug 27th 2009, 05:51 PM
chengbin
Quote:

Originally Posted by xwrathbringerx
How do I exactly differentiate 3^(5x)

Thanxx a lot!

$\displaystyle a^x=(x')a^x \ln a$
• Aug 27th 2009, 05:51 PM
Calculus26
The general fromula for f= a^(u(x))

f ' = ln(a) a^(u(x))du/dx

For 3^(5x)

f ' = ln(3)3^(5x)5 = 5ln(3)3^(5x)
• Aug 27th 2009, 06:04 PM
Calculus26
http://www.mathhelpforum.com/math-he...04062c89-1.gif

doesn't make sense as x ' = 1

(a^x ) ' = ln(a)*a^x period

for a^(u(x) use the chain rule to obtain

f ' = ln(a) a^(u(x))du/dx
• Aug 28th 2009, 03:16 AM
HallsofIvy
A general method for problems like this is to take the logarithm of both sides. If $\displaystyle y= 3^{5x}$ then $\displaystyle ln(y)= ln(3^{5x})= 5x ln(3)$. Now differentiate both sides with respect to x.

The derivative of ln(y) with respect to y is $\displaystyle \frac{1}{y}$ but since y itself is a function of x, by the chain rule, the derivative of ln(y) with respect to x is $\displaystyle \frac{1}{y}\frac{dy}{dx}$.

$\displaystyle \frac{1}{y}\frac{dy}{dx}= 5 ln(3)$ so $\displaystyle \frac{dy}{dx}= 5 ln(3) y= 5 ln(3) 3^{5y}$.