## Calculus: Early Transcendentals (2nd Edition)

$(a)$. $$\frac{dy}{dx}=\frac{-3}{2y}$$ $(b)$. The slope of the tangent line to the curve is $m=\frac{-3\sqrt{5}}{10}$
$(a)$ Use implicit differentiation to determine $\frac{dy}{dx}$ for $y^2+3x=8$ Taking the derivative implicitly we get: $$2y\frac{dy}{dx}+3=0$$ solve for $\frac{dy}{dx}$ $$\frac{dy}{dx}=\frac{-3}{2y}$$ $(b)$. Find the slope of the tangent line to the curve at $(1,\sqrt{5})$ We plug in the point $(1,\sqrt{5})$ into the derivative from part $(a)$. Hence: $$\frac{dy}{dx}=\frac{-3}{2\sqrt{5}}=\frac{-3\sqrt{5}}{10}$$