## Calculus 10th Edition

$y=\dfrac{-3x+25}{4}.$
Using the Chain Rule: $u=25-x^2$; $\dfrac{du}{dx}=-2x$ $f(u)=\sqrt{u};\dfrac{d}{du}f(u)=\dfrac{1}{2\sqrt{u}}$ $\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=-\dfrac{x}{\sqrt{25-x^2}}$ $f'(3)=-\dfrac{3}{\sqrt{25-3^2}}=-\dfrac{3}{4}.$ Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y-4)=-\dfrac{3}{4}(x-3)\rightarrow y=\dfrac{-3x+25}{4}.$ A graphing utility is used to graph the function f and the specified tangent line. The intersection point at (3, 4) is confirmed.