Suppose $y=f(x)$ to find $\displaystyle{\frac{d}{dx} 4y^3}$.
Suppose $y=f(x)$ to find $\displaystyle{\frac{d}{dx} x^5 y^3}$.
Suppose $x=g(t)$ to find $\displaystyle{\frac{d}{dt} (t + \cos x^3)}$.
Find $\displaystyle{\frac{dy}{dx}}$ and $\displaystyle{\frac{d^2y}{dx^2}}$ if $x^2 + y^2 = r^2$ for some constant $r$.
First we find $\displaystyle{\frac{dy}{dx} = \frac{-x}{y}}$. Then, upon differentiating a second time and substituting in for the first derivative, we obtain $\displaystyle{\frac{d^2 y}{dx^2} = -\frac{x^2+y^2}{y^3}}$.
Find the equation of the tangent line at $(1,1)$ to the graph of $x^2y + 3y^4 = x^3 y^3 + 3$.
Find $\displaystyle{\frac{d^2y}{dx^2}}$ if $x^2 + 3y^2 = xy + 3$.
First, we find $\displaystyle{\frac{dy}{dx} = \frac{y-2x}{6y-x}}$. Then, upon differentiating again, simplifying, and substituting the first derivative, we obtain $\displaystyle{\frac{d^2 y}{dx^2} = -\frac{11(x^2 - xy + 6y)}{(x - 6y)^2}}$.