Find the equation of the tangent to the curve $y = {2^x} + {2^{ - x}}$ at the point $(2,4{1 \over 4})$ $\eqalign{
  & y = {2^x} + {2^{ - x}}  \cr 
  & \ln y = x\ln 2 - x\ln 2  \cr 
  & \ln y = 0  \cr 
  & {1 \over y}{{dy} \over {dx}} = 0  \cr 
  & {{dy} \over {dx}} = 0 \cr} $
I've checked the answer and I've got the differential wrong, What am I doing wrong? I assume it has something to do with the expression with the negative exponent? Am I not allowed to prefix the natural expression with "minus" x? Could someone explain please, thank you!

$$y=2^x+2^{-x}\implies y'=2^x\log2-2^{-x}\log 2=\left(2^x-2^{-x}\right)\log 2$$

Neat off-topic stuff: consider $y=b^x+b^{-x}$ for some positive real $b$. Notice that differentiating yields $y'=(b^x-b^{-x})\log b$. In the case where $b=e$, we have $y=e^x+e^{-x}$ as well as $y'=e^x-e^{-x}$. Taking the second derivative gives us $y''=e^x+e^{-x}=y$ so we've found a solution to $y''-y=0$. What about $y=e^{ix}+e^{-ix}$? Isn't this a solution to $y''+y=0$? What other solutions to $y''-y=0$ do you know?
