Activity 35.
For each of the following curves, use implicit differentiation to find \(dy/dx\) and determine the equation of the tangent line at the given point.
(a)
\(x^3 - y^3 = 6xy\text{,}\) Â \((-3,3)\)
Hint.
Note that \(\frac{d}{dx}[6xy]\) requires the product rule.
Answer.
\(\frac{dy}{dx}(-3y^2 - 6x) = 6y-3x^2 \) and the tangent line has equation \(y - 3 = 1(x+3)\text{.}\)
Solution.
Differentiating with respect to \(x\text{,}\)
\begin{equation*}
\frac{d}{dx}[x^3 - y^3] = \frac{d}{dx}[6xy]\text{,}
\end{equation*}
so that by the chain and product rules we have
\begin{equation*}
3x^2 - 3y^2 \frac{dy}{dx} = 6x\frac{dy}{dx}+ 6y\text{.}
\end{equation*}
Rearranging to get all terms with \(\frac{dy}{dx}\) on the same side, it follows that
\begin{equation*}
-3y^2 \frac{dy}{dx} - 6x\frac{dy}{dx} = 6y-3x^2\text{,}
\end{equation*}
and thus
\begin{equation*}
\frac{dy}{dx}(-3y^2 - 6x) = 6y-3x^2\text{.}
\end{equation*}
Finally, we have established that
\begin{equation*}
\frac{dy}{dx} = \frac{6y-3x^2}{-3y^2 - 6x}\text{,}
\end{equation*}
so evaluating at \((-3,3)\text{,}\) we have \(\left. \frac{dy}{dx} \right|_{(-3,3)} = \frac{6(3)-3(-3)^2}{-3(3)^2 - 6(-3)} = 1\text{.}\) Thus, the tangent line has equation \(y - 3 = 1(x+3)\text{.}\)
(b)
\(\sin(y) + y = x^3 + x\text{,}\) Â \((0,0)\)
Hint.
With \(y\) being a function of \(x\text{,}\) \(\frac{d}{dx}[\sin(y)]\) requires the chain rule.
Answer.
\(\frac{dy}{dx} = \frac{3x^2 + 1}{\cos(y) + 1}\) and the tangent line has equation \(y = \frac{1}{2}x\text{.}\)
Solution.
After differentiating with respect to \(x\text{,}\) we have
\begin{equation*}
\cos(y) \frac{dy}{dx} + \frac{dy}{dx} = 3x^2 + 1\text{.}
\end{equation*}
Taking the usual steps to solve for \(\frac{dy}{dx}\text{,}\) we find that
\begin{equation*}
\frac{dy}{dx} = \frac{3x^2 + 1}{\cos(y) + 1}\text{.}
\end{equation*}
Evaluating the slope of the tangent line at \((0,0)\text{,}\) we have \(\left. \frac{dy}{dx} \right|_{(0,0)} = \frac{1}{2}\text{,}\) and thus the tangent line at \((0,0)\) has equation \(y = \frac{1}{2}x\text{.}\)
(c)
\(3x e^{-xy} = y^2\text{,}\) Â \((0.619061,1)\)
Hint.
To calculate \(\frac{d}{dx}[x e^{-xy}]\text{,}\) first use the product rule and temporarily defer computing \(\frac{d}{dx}[e^{-xy}]\text{.}\)
Answer.
\(\frac{dy}{dx} = \frac{3e^{-xy} - 3xye^{-xy}}{3x^2e^{-xy}+2y}\) and the tangent line is \(y - 1 = 0.234950(x - 0.619061)\text{.}\)
Solution.
When we differentiate both sides with respect to \(x\text{,}\)
\begin{equation*}
\frac{d}{dx}[3x e^{-xy}] = \frac{d}{dx}[y^2]\text{,}
\end{equation*}
we first observe that the product rule is needed on the left and the chain rule on the right. Applying those rules, we have
\begin{equation*}
3x\frac{d}{dx}[e^{-xy}] + 3e^{-xy} = 2y\frac{dy}{dx}\text{.}
\end{equation*}
Next, we apply the chain rule to differentiate \(e^{-xy}\text{,}\) which yields
\begin{equation*}
3xe^{-xy}\frac{d}{dx}[-xy] + 3e^{-xy} = 2y\frac{dy}{dx}\text{.}
\end{equation*}
Finally, to complete the process of differentiation, we use the product rule and get
\begin{equation*}
3xe^{-xy}(-x\frac{dy}{dx} - y) + 3e^{-xy} = 2y\frac{dy}{dx}\text{.}
\end{equation*}
To solve for \(\frac{dy}{dx}\text{,}\) we first expand to have
\begin{equation*}
-3x^2e^{-xy}\frac{dy}{dx} - 3xye^{-xy} + 3e^{-xy} = 2y\frac{dy}{dx}\text{,}
\end{equation*}
and then the usual algebraic work may be done to deduce that
\begin{equation*}
\frac{dy}{dx} = \frac{3e^{-xy} - 3xye^{-xy}}{3x^2e^{-xy}+2y}\text{.}
\end{equation*}
Evaluating at the point \((0.619061,1)\text{,}\) it follows that the slope of the tangent line is
\begin{equation*}
\left. \frac{dy}{dx} \right|_{(0.619061,1)} = \frac{3 e^{-0.619061} - 3 \cdot 0.619061 e^{-0.619061}}{3(0.619061)^2e^{-0.619061}+2} \approx 0.234950\text{.}
\end{equation*}
Thus, the tangent line is given by \(y - 1 = 0.234950(x - 0.619061)\text{.}\)
