How do I exactly differentiate 3^(5x)

Thanxx a lot!

Printable View

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

Thanxx a lot! - August 27th 2009, 05:51 PMchengbin
- August 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) - August 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 - August 28th 2009, 03:16 AMHallsofIvy
A general method for problems like this is to take the logarithm of both sides. If then . Now differentiate both sides

**with respect to x**.

The derivative of ln(y)**with respect to y**is but since y itself is a function of x, by the chain rule, the derivative of ln(y) with respect to x is .

so .