Question

How do you find the derivative of natural logs?
Find the derivative h(z)=(ln9)^z
Thanks.

Answer 1

I assume you want to take the derivative with respect to z.

use the idea
\[ e^{\ln(a)}= a\]
in this case, let a= ln(9)
so
\[ \ln(9) = e^{\ln( \ln(9))}\]
rewrite your expression as
\[ \left( e^{\ln( \ln(9))}\right)^z = e^{\ln( \ln(9)) z}\]
now that might look ugly, but ln(ln(9)) is just an (ugly) constant number

so you are taking the derivative of
\[ \frac{ d\ e^{bz}}{dz} \]
where b is a constant= ln(ln(9))

Answer 2

Thanks, but I'm not getting the correct answer when I input e^ln(ln(9))z

Answer 3

that's because the
\[ \frac{d}{dz} e^{bz} = e^{bz} \frac{d}{dz}bz = b e^{bz} \]

Answer 4

so the answer is
\[ \ln(\ln(9)) e^{\ln(\ln(9))z} \]

Answer 5

they may re-write e^(ln(ln(9)) as ln(9)
so another way to write the answer is
\[ \ln(\ln(9)) (\ln(9))^z \]

Answer 6

Oohhh thank you! That makes much more sense now. But why isn't z taken down as well?

Answer 7

\[ \frac{d}{dx} e^x = e^x \]

Khan goes into detail about this if you want more info.

http://www.khanacademy.org/math/calculus/differential-calculus/product_rule/v/derivatives-of-sin-x--cos-x--tan-x--e-x-and-ln-x

http://www.khanacademy.org/math/calculus/differential-calculus/der_common_functions/v/proof--d-dx-e-x----e-x