evaluating derivative of $\log_4(2x^2+1)$ 
Find the derivative and evaluate at $f\;'(2):$ $$\log_4(2x^2+1)$$  

$\log_4(2x^2+1)=y$
$4^y=2x^2+1$  
$4^y\ln4 \times y\;'=4x$
$y\;'=\dfrac{4x}{4^y\ln4}\implies \dfrac{4x}{(2x^2+1)\ln4}$  
What am I doing wrong? I evaluated at $2$ and got $1.154$

There seem to be no issues with your derivation; when I plug in x=2 I get 0.641 to three decimal places...perhaps you substituted in the wrong value? 
