Can somebody help me find the derivative of :
m = 2(2)^(-t/15)
when t = 5...
please and thank you
you have $\displaystyle f(t)=2\cdot{2^{\frac{-t}{15}}=2^{\frac{-t}[15}+1}$ so if we take the ln both sides we get $\displaystyle ln(f(t))=\bigg[\frac{-t}{15}+1\bigg]\cdot{ln(2)}$ so by taking the derivative of each side we get $\displaystyle \frac{f'(t)}{f(t)}=\frac{-ln(2)}{15}$ and since $\displaystyle f(t)$ is our original equation we have $\displaystyle f'(t)=\frac{-ln(2)2^{\frac{-t}{15}+1}}{15}$