Answer
$y=\dfrac{13x-8}{9}.$
Work Step by Step
$f(x)=\frac{1}{3}x\sqrt{x^2+5}=\frac{1}{3}\sqrt{(x^2)(x^2+5)}=\frac{1}{3}\sqrt{x^4+5x^2}.$
$u=x^4+5x^2$; $\dfrac{du}{dx}=4x^3+10x.$
$\dfrac{d}{du}f(u)=\dfrac{1}{6\sqrt{u}}$
$\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=\dfrac{2x^3+5x}{3\sqrt{x^4+5x^2}}=\dfrac{2x^2+5}{3\sqrt{x^2+5}}.$
$f'(2)=\dfrac{2(2)^2+5}{3\sqrt{2^2+5}}=\dfrac{13}{9}.$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-2)=\dfrac{13}{9}(x-2)\rightarrow y=\dfrac{13x-8}{9}.$
A graphing calculator and a computer algebra system have been used to confirm these results.
