How do I exactly differentiate 3^(5x)

Thanxx a lot!

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- Aug 27th 2009, 05:38 PMxwrathbringerxSimple Differentiation
How do I exactly differentiate 3^(5x)

Thanxx a lot! - Aug 27th 2009, 05:51 PMchengbin
- Aug 27th 2009, 05:51 PMCalculus26
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 PMCalculus26
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 AMHallsofIvy
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 $\displaystyle \frac{dy}{dx}= 5 ln(3) y= 5 ln(3) 3^{5y}$.