1. ## derivatives of log/ln

Can somebody help me find the derivative of :

m = 2(2)^(-t/15)

when t = 5...

2. ## ok

Originally Posted by go.leafs.go.89
Can somebody help me find the derivative of :

m = 2(2)^(-t/15)

when t = 5...

you have $f(t)=2\cdot{2^{\frac{-t}{15}}=2^{\frac{-t}[15}+1}$ so if we take the ln both sides we get $ln(f(t))=\bigg[\frac{-t}{15}+1\bigg]\cdot{ln(2)}$ so by taking the derivative of each side we get $\frac{f'(t)}{f(t)}=\frac{-ln(2)}{15}$ and since $f(t)$ is our original equation we have $f'(t)=\frac{-ln(2)2^{\frac{-t}{15}+1}}{15}$

3. Originally Posted by go.leafs.go.89
Can somebody help me find the derivative of :

m = 2(2)^(-t/15)

when t = 5...

Try expressing the two raised to a power as e raised to a power:

$m = 2~2^{-t/15}=2~e^{- \frac{t~\ln(2)}{15}}$

RonL

4. hm i dont understand how the plus 1 came?
and also .. we didn't learn about implicit differentiation ..
is there any other way to do this question?

5. ## Haha

Originally Posted by go.leafs.go.89
hm i dont understand how the plus 1 came?
and also .. we didn't learn about implicit differentiation ..
is there any other way to do this question?
You know the name of something you never learned? ok...but still the +1 came from the fact taht $2\cdot{2^{n}}=2^{n+1}$ and you can do what CaptainBlack said and use the rules of $\frac{D[e^{u(x)}]}{dx}$