Section 2.6 Implicit Differentiation
Sometimes we have an equation that can't easily be solved for \(y\text{,}\) for example: \(x^3 + 3y^3 = 3xy\text{.}\)
It helps to consider \(y = y(x)\text{.}\) With that consideration, it might be possible to calculate \(\dfrac{dy}{dx}\text{.}\)
Differentiate \(x^3 + 3y^3 = 3xy\text{:}\)
\begin{align*}
\dfrac{d}{dx}(x^3 + 3y^3) \amp = \dfrac{d}{dx}(3xy)\\
\dfrac{d}{dx}(x^3) + \dfrac{d}{dx}(3y^3) \amp = 3x \cdot \dfrac{dy}{dx} + y \cdot \dfrac{d}{dx}(3x) \amp \text{(separate the terms)}\\
3x^2 + 3 \cdot 3y^2 \dfrac{dy}{dx} \amp = 3x \cdot \dfrac{dy}{dx} + y \cdot 3 \amp \text{(apply the chain rule)}\\
3x^2 + 9y^2 \dfrac{dy}{dx} \amp = 3x \dfrac{dy}{dx} + 3y \amp \text{(solve for }\dfrac{dy}{dx})\\
9y^2 \dfrac{dy}{dx} - 3x\dfrac{dy}{dx} \amp = 3y - 3x\\
\dfrac{dy}{dx}(9y^2 -3x) \amp = 3y - 3x\\
\dfrac{dy}{dx} \amp = \dfrac{3y - 3x}{9y^2 -3x}
\end{align*}
Example 2.6.1.
Find the slope of the tangent to \(x^2 + y^2 = 1\text{,}\) at the point \(\left( \dfrac{1}{2},\dfrac{\sqrt{3}}{2} \right)\text{:}\)
\begin{align*}
x^2 + (y(x))^2 \amp = 1 \amp \text{(treat y as y(x))}\\
\dfrac{d}{dx} \left(x^2 + (y(x))^2 \right) \amp = \dfrac{d}{dx}(1)\\
2x + 2y \cdot \dfrac{dy}{dx} \amp = 0\\
2y \cdot \dfrac{dy}{dx} \amp = -2x\\
\dfrac{dy}{dx} \amp = - \dfrac{x}{y}
\end{align*}
At \(\left( \dfrac{1}{2},\dfrac{\sqrt{3}}{2} \right)\text{,}\) \(M_{\text{tan}} = \dfrac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\dfrac{1}{\sqrt{3}}\)