We put

u = x^{2} + 1,

giving

u' = 2x

Applying the formula, we have:

`(dy)/(dx)=3\ (log_7e)(2x)/(x^2+1)` `=6\ (log_7e) x/(x^2+1)` `=3.083 x/(x^2+1)`

`(dy)/(dx)=3\ (log_7e)(2x)/(x^2+1)`

`=6\ (log_7e) x/(x^2+1)`

`=3.083 x/(x^2+1)`

The value 3.083 comes from using the change of base rule.