Answer
The equation of the tangent line is $y=\dfrac{2x+16}{25}.$
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=4x; u'(x)=4$
$v(x)=x^2+6; v'(x)=2x$
$f'(x)=\frac{4(x^2+6)-(4x)(2x)}{(x^2+6)^2}=\frac{24-4x^2}{(x^2+6)^2}$
$f'(2)=\frac{24-4(2)^2}{((2^2)+6)^2}=\frac{2}{25}$.
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-\frac{4}{5})=\frac{2}{25}(x-2) \rightarrow y=\frac{2x+16}{25}.$
