Differentiate log(base 7)(4x^2+1)
Use the change of base rule to turn it into a natural logarithm function.
$\displaystyle y = \log_7{(4x^2 + 1)}$
$\displaystyle = \frac{\ln{(4x^2 + 1)}}{\ln{7}}$
$\displaystyle = \frac{1}{\ln{7}}\cdot\ln{(4x^2 + 1)}$.
Let $\displaystyle u = 4x^2 + 1$ so that $\displaystyle y= \frac{1}{\ln{7}}\cdot\ln{u}$.
$\displaystyle \frac{du}{dx} = 8x$
$\displaystyle \frac{dy}{du} = \frac{1}{\ln{7}}\cdot\frac{1}{u}$
$\displaystyle = \frac{1}{(4x^2 + 1)\ln{7}}$.
Therefore
$\displaystyle \frac{dy}{dx} = \frac{8x}{(4x^2 + 1)\ln{7}}$.