let f(x)=log8(4x^2 + 1). then, f(2)=

Explanation:

Step1: Recall derivative formula for $\log_a(u)$

The derivative of $y = \log_a(u)$ is $y'=\frac{1}{u\ln(a)}\cdot u'$. Here $a = 8$ and $u=4x^{2}+1$.

Step2: Find the derivative of $u$

If $u = 4x^{2}+1$, then $u'=8x$ using the power - rule $(x^n)'=nx^{n - 1}$.

Step3: Find the derivative of $f(x)$

$f'(x)=\frac{1}{(4x^{2}+1)\ln(8)}\cdot8x=\frac{8x}{(4x^{2}+1)\ln(8)}$.

Step4: Evaluate $f'(2)$

Substitute $x = 2$ into $f'(x)$. We have $f'(2)=\frac{8\times2}{(4\times2^{2}+1)\ln(8)}=\frac{16}{(16 + 1)\ln(8)}=\frac{16}{17\ln(8)}$.

Answer:

$\frac{16}{17\ln(8)}$