Find $\frac{d y}{d x}$ by implicit differentiation for the following equation. \[ e^{x^{3} y}=5 x+6 y+4 \] \[ \frac{\mathrm{dy}}{\mathrm{dx}}=\square \]
Solution
Step 1 ：Differentiate both sides of the equation with respect to \(x\). We get \(e^{x^{3} y} \cdot \frac{d}{dx}(x^{3} y) = 5 + 6 \cdot \frac{dy}{dx}\).
Step 2 ：Differentiate \(x^{3} y\) with respect to \(x\) using the product rule. We get \(3x^{2}y + x^{3} \cdot \frac{dy}{dx}\). Substitute this into the equation from step 1, we get \(e^{x^{3} y} \cdot (3x^{2}y + x^{3} \cdot \frac{dy}{dx}) = 5 + 6 \cdot \frac{dy}{dx}\).
Step 3 ：Rearrange the equation to solve for \(\frac{dy}{dx}\). First, distribute \(e^{x^{3} y}\) on the left side of the equation, we get \(3x^{2}y \cdot e^{x^{3} y} + x^{3} \cdot e^{x^{3} y} \cdot \frac{dy}{dx} = 5 + 6 \cdot \frac{dy}{dx}\). Then, move the terms involving \(\frac{dy}{dx}\) to one side and the rest to the other side, we get \(x^{3} \cdot e^{x^{3} y} \cdot \frac{dy}{dx} - 6 \cdot \frac{dy}{dx} = 5 - 3x^{2}y \cdot e^{x^{3} y}\). Finally, factor out \(\frac{dy}{dx}\) and divide both sides by the remaining factor on the left side, we get \(\frac{dy}{dx} = \frac{5 - 3x^{2}y \cdot e^{x^{3} y}}{x^{3} \cdot e^{x^{3} y} - 6}\).
Step 4 ：\(\boxed{\frac{dy}{dx} = \frac{5 - 3x^{2}y \cdot e^{x^{3} y}}{x^{3} \cdot e^{x^{3} y} - 6}}\) is the solution.
