Explicit equation (i.e., \(y\) given explicitly as a function of \(x\)):
$$y=e^{3x}-x^2+\ln x$$
Explicit equation (i.e., \(y\) given explicitly as a function of \(x\), or \(x\) given explicitly as a function of \(y\)):
$$x^2y+y^2=e^{xy}+\sin(x+y)$$
For functions given IMPLICITLY:
Step 1. Assuming that \(y\) is a function of \(x\), i.e., \(y = f (x)\), take the usual derivative with respect to \(x\), keeping in mind to use the chain rule when differentiating terms involving \(y\) (which, of course, involves multiplying by \(y^{\prime}\text{or }\frac{dy}{dx}).\)
Step 2. Solve for \(y'\), if possible, or requested.
Example
Differentiate the following equations.
$$\begin{array}{cc}(\text{a})&&2x^3-3xy^2+y^4=-20\\\\\text{(b)}&&xy=e^{-xy}\end{array}$$
Solution
(a) Step 1.
$$\begin{aligned}
2x^{3}-3xy^{2}+y^{4}& =-20 \\
6x^{2}-3[y^{2}+2yy^{\prime}x]+4y^{3}y^{\prime}& =0 \\
6x^{2}-3y^{2}-6yy^{\prime}x+4y^{3}y& =0
\end{aligned}$$
Step 2.
$$\begin{aligned}
-6xyy+4y^{3}y^{\prime}& =-6x^{2}+3y^{2} \\
y'(-6xy+4y^{3})& =-6x^{2}+3y^{2}, \\
\mathcal{y}^{\prime}& =\frac{-6x^{2}+3y^{2}}{-6xy+4y^{3}}
\end{aligned}$$
(b) Step 1.
$$\begin{aligned}
\boldsymbol{x}y& =e^{-xy} \\
y+xy& =e^{-xy}(-y-xy^{\prime}) \\
&=-e^{-xy}y-e^{-xy}xy^{\prime}
\end{aligned}$$
Step 2.
$$\begin{aligned}
xy^{\prime}+e^{-xy}xy& =-e^{-xy}y-y \\
y^{\prime}(x+e^{-xy}x)& =-y(e^{-xy}+1)\quad, \\
\mathcal{y}^{\prime}& =\frac{-y(e^{-xy}+1)}{x(e^{-xy}+1)}=\frac{-y}{x} \\
\mathrm{since~}e^{-xy}+1& \neq\mathbf{0}.
\end{aligned}$$