## Calculus 10th Edition

$y=\dfrac{14-2x}{3}.$
Using the Chain Rule: $u=(9-x^2)$; $\dfrac{du}{dx}=-2x$ $\dfrac{d}{du}f(u)=\dfrac{2}{3\sqrt[3]{u}}$ $\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=-\dfrac{4x}{3\sqrt[3]{9-x^2}}.$ $f'(1)=-\dfrac{4(1)}{3\sqrt[3]{9-1^2}}=-\dfrac{2}{3}.$ Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y-4)=-\dfrac{2}{3}(x-1)\rightarrow y=\dfrac{14-2x}{3}.$ A graphing calculator and a computer algebra system have been used to confirm these results.