Feb 28, 2014 Thread starter #1 T tmt Active member Jan 15, 2014 236 Hi, In my notes I put down that y=2^x y'=2^x ln(x) However, I seem to remember that it is in fact y'=2^x ln(2) Which one is correct? Thanks, Tim

Hi, In my notes I put down that y=2^x y'=2^x ln(x) However, I seem to remember that it is in fact y'=2^x ln(2) Which one is correct? Thanks, Tim

Feb 28, 2014 Admin #2 Ackbach Indicium Physicus Staff member Jan 26, 2012 4,202 \begin{align*} y&=2^x \\ \ln(y)&= \ln(2^x) \\ \ln(y)&= x \ln(2) \\ \frac{y'}{y}&=\ln(2) \\ y'&= y \ln(2) \\ y'&= \ln(2) \, 2^x. \end{align*}

\begin{align*} y&=2^x \\ \ln(y)&= \ln(2^x) \\ \ln(y)&= x \ln(2) \\ \frac{y'}{y}&=\ln(2) \\ y'&= y \ln(2) \\ y'&= \ln(2) \, 2^x. \end{align*}

Feb 28, 2014 Admin #3 M MarkFL Administrator Staff member Feb 24, 2012 13,775 Another way to view it (although I favor Ackbach's method) is: \(\displaystyle y=2^x=e^{\ln\left(2^x \right)}=e^{x\ln(2)}\) And so: \(\displaystyle y'=e^{x\ln(2)}\cdot\ln(2)=\ln(2)2^x\)

Another way to view it (although I favor Ackbach's method) is: \(\displaystyle y=2^x=e^{\ln\left(2^x \right)}=e^{x\ln(2)}\) And so: \(\displaystyle y'=e^{x\ln(2)}\cdot\ln(2)=\ln(2)2^x\)