음함수의 미분법

First, we compute $\frac{dy}{dx}$ by implicit differentiation. Differentiating both sides of the given equation with respect to $x$ while treating $y$ as a function of $x$, we see that

$$6y^2 \frac{dy}{dx} + 2y \frac{dy}{dx} - 5y^4 \frac{dy}{dx} = 4x^3 - 6x^2 + 2x\,.$$ Removing a factor of $\frac{dy}{dx}$ from the lefthand side and dividing to solve for $\frac{dy}{dx}$, we have

$$\frac{dy}{dx} = \frac{4x^3 - 6x^2 + 2x}{6y^2 + 2y - 5y^4} = \frac{2x(2x^2 - 3x + 1)}{y(-5y^3 + 6y + 2)}\,.$$ The tangent line is potentially horizontal where the numerator of $\frac{dy}{dx}$ is zero and potentially vertical where $\frac{dy}{dx}$ is zero. Solving $2x(2x^2 - 3x + 1) = 0$, we see that

$$2x(2x-1)(x-1) = 0$$

so the numerator is zero at $x = 0, 0.5, 1$. The denominator is zero where $y(-5y^3 + 6y + 2) = 0$. Using technology to estimate where $-5y^3 + 6y + 2 = 0$, we see the denominator is zero at $y = 0$ and $y \approx -0.856, -0.379, 1.235$.

Next, we need to find the coordinates of these points where potential horizontal or vertical tangent lines occur. First, we consider the potential horizontal tangent line locations. When $x = 0$, $2y^3+y^2-y^5 = x^4 - 2x^3 + x^2 = 0$, and thus we know $y^2(2y + 1 - y^3) = 0$, from which it follows that $y = 0$, $y=-1$ or $y \approx -0.618, 1.618$. This leads to four points to consider: $(0,0)$, $(0,-1)$, $(0,-0.618)$, and $(0,1.618)$. It is apparent from the graph shown below that the tangent line at $(0,0)$ is not horizontal (here both the numerator and denominator of $\frac{dy}{dx}$ are zero), but at each of the other three points, the tangent line is horizontal.

We now reason similarly for the other $x$-values where the numerator is zero. At $x = 1$, $2y^3+y^2-y^5 = 1^4 - 2\cdot 1^3 + 1^2 = 0$, and here again $y = 0$, $y=-1$ or $y \approx -0.618, 1.618$. This leads to four potential points where the tangent line is horizontal: $(1,0)$, $(1,-1)$, $(1,-0.618)$, and $(1,1.618)$. At $(1,0)$, there's not a horizontal tangent line, but at each of the others there is, as seen in the figure. Similar reasoning when $x = 0.5$ shows that $2y^3+y^2-y^5 = 0.5^4 - 2\cdot 0.5^3 + 0.5^2 = 0.625$, from which we find that $y \approx -1.0493, 0.2104, 1.6139$, and thus the three points $(0.5,-1.0493)$, $(0.5,0.2104)$, $(0.5, 1.6139)$.

Returning to where the denominator is zero ($y = 0$ and $y \approx -0.856, -0.379, 1.235$), we now determine points where the tangent line is vertical. Is is apparent from the figure below that there are no vertical tangent lines when $y = 0$ or $y \approx -0.856$, so there are two values to pursue: $y \approx -0.379, 1.235$.

When $y \approx -0.379$, it follows that $2(-0.379)^3+(-0.379)^2-(-0.379)^5 = x^4 - 2x^3 + x^2$, so

$$x^4 - 2x^3 + x^2 = 0.0426$$

from which we find that $x \approx -0.1756, 0.2912, 0.7088, 1.1756$. The figure confirms that there are vertical tangent lines at $(-0.1756,-0.379)$, $(0.2912,-0.379)$, $(0.7088,-0.379)$, and $(1.1756,-0.379)$

Lastly, when $y \approx 1.235$, it follows $2(1.235)^3+(1.235)^2-(1.235)^5 = x^4 - 2x^3 + x^2$, so

$$x^4 - 2x^3 + x^2 = 2.4195$$

and thus via technology, $x \approx -0.8437, 1.8437$. This leads to two additional points at which the tangent line is vertical: $(-0.8437, 1.235)$ and $(1.8437, 1.235)$.

All of our above work is summarized in the following figure.