evaluating derivative of $\log_4(2x^2+1)$

Question

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 $1.154$

Answer 1

Notice that $(\log_ax)'=\frac{1}{x\ln a}$. SO

$$(\log_4(2x^2+1))\big|_{x=2}=\frac{1}{(2x^2+1)\ln 4} \times 4x\big|_{x=2}=\frac{4}{9\ln 2}$$

Answer 2

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?

Answer 3

You result $$y'=\dfrac{4x}{(2x^2+1)\ln4}$$ is perfectly correct. So, if $x=2$, $y'=\frac{8}{9 \log (4)}=0.641198$