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!

$$\ln(2^x+2^{-x}) \neq \ln(2^x)+\log(2^{-x})$$
Recall that $\ln(a+b) \neq \ln(a)+\ln(b)$. We do have that $\ln (ab) = \ln a + \ln b$
$$y=2^x+2^{-x}$$
$$ y'=\dfrac{d}{dx}(2^x) + \dfrac{d}{dx}(2^{-x})\tag{1}$$ 
$$y' = 2^x \ln 2 + (-1)\cdot 2^{-x}\ln 2 \tag{2}$$
$$y' = 2^x \ln 2 - 2^{-x}\ln 2 \tag{3}$$ 
$$ y'= \left(2^x-2^{-x}\right)\ln 2\tag{4}$$
$(2)$ In general, if you have $$y = a^x;\;\; \text{then}\;\; y' =\dfrac{dy}{dx} = a^x\ln a\tag{a is constant}$$
$(3)$ Similarly, if you have $$y = a^{f(x)},\;\; \text{then}\;\; y'= \dfrac{dy}{dx} = a^{f(x)}f'(x)(\ln a)\qquad\tag{a is constant}$$
