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Derivative of 2^x

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
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
\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*}
 

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:

\(\displaystyle y'=e^{x\ln(2)}\cdot\ln(2)=\ln(2)2^x\)