## Calculus (3rd Edition)

a) $\frac{dy}{dt}=(\frac{x^{2}}{y^{2}})\frac{dx}{dt}$ b) $\frac{dy}{dt}=-(\frac{y+x}{2y^{3}+x})\frac{dx}{dt}$
a) Implicitly differentiating $x^{3}-y^{3}=1$ with respect to $t$, we have $3x^{2}\frac{dx}{dt}-3y^{2}\frac{dy}{dt}=0$ $\implies 3y^{2}\frac{dy}{dt}=3x^{2}\frac{dx}{dt}$ $\implies \frac{dy}{dt}=\frac{3x^{2}}{3y^{2}}\frac{dx}{dt}=(\frac{x^{2}}{y^{2}})\frac{dx}{dt}$ b) Implicitly differentiating $y^{4}+2xy+x^{2}=0$ with respect to $t$, we have $4y^{3}\frac{dy}{dt}+2(\frac{dx}{dt}\times y+x\frac{dy}{dt})+2x\frac{dx}{dt}=0$ $\implies 4y^{3}\frac{dy}{dt}+2x\frac{dy}{dt}=-2y\frac{dx}{dt}-2x\frac{dx}{dt}$ $\implies \frac{dy}{dt}(4y^{3}+2x)=(-2y-2x)\frac{dx}{dt}$ $\implies \frac{dy}{dt}=(\frac{-2y-2x}{4y^{3}+2x})\frac{dx}{dt}=(\frac{-y-x}{2y^{3}+x})\frac{dx}{dt}$ $=-(\frac{y+x}{2y^{3}+x})\frac{dx}{dt}$