## Calculus: Early Transcendentals (2nd Edition)

$(a)$. $$\frac{dy}{dx}=\frac{-x^3}{y^3}$$ $(b)$. Slope of tangent line is $m=1$
$(a).$ Use implicit differentiation to find $\frac{dy}{dx}$ for $x^4+y^4=2$ Taking the derivative implicitly for the above function we get: $$4x^3+4y^3\frac{dy}{dx}=0$$ Solve for $\frac{dy}{dx}$ $$\frac{dy}{dx}=\frac{-x^3}{y^3}$$ $(b).$ Find the slope of the curve tangent to the point $(1,-1)$ for the above function The slope of our tangent line is determined by plugging in $(1,-1)$ into the derivative from part $(a)$. Hence: $$\frac{dy}{dx}=\frac{-(1)^3}{(-1)^3}=\frac{-1}{-1}=1$$