Finding dx dy

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In this section we differentiate equations without expressing them in terms of a single variable.

Turn it around

When we see an equation, like $x + y^3 - y = 1$ , $x$ and $y$ are both quantities that change according to that formula. We typically consider $y$ as a function of $x$ , but we can also consider $x$ as a function of $y$ .

The equation $x + y^3 - y = 1$ defines $x$ implicitly as a function of $y$ . Find $\dd [x]{y}$ .

$\begin{align*} \dd{y} (x + y^3- y ) &= \dd{y} (1)\\ \dd[x]{y} + 3y^2 - 1 &= 0\\ \dd[x]{y} = 1 - 3y^2 \end{align*}$

We know that at points with $\dd [x]{y} = 0$ , our graph has horizontal tangent lines. Something similar can be said for vertical tangent lines.

Finding Vertical Tangent Lines If a point on the graph of an equation has $\dd [y]{x}= 0$ , then the tangent line at that point is vertical.

Find the points on the graph of $x + y^3 - y = 1$ where the tangent line is vertical.

We have already seen that $\dd [x]{y}= 1-3y^2$ . $\begin{align*} \dd[x]{y} &= 0\\ 1 -3y^2 &= 0\\ 3y^2 &= 1\\ y^2 &= \answer{\frac{1}{3}}\\ y &= \pm \answer{\frac{1}{\sqrt{3}}} \end{align*}$

Plugging these back into the equation:

$\begin{align*} y &= \frac{1}{\sqrt{3}}\\ x &= 1 + y - y^3\\ &= 1 + \frac{2}{3\sqrt{3}} \end{align*}$

One point is $\left ( 1 + \frac {2}{3\sqrt {3}} , \frac {1}{\sqrt {3}} \right )$ . Plugging in the other value of $y$ gives the other point: $\left ( 1 - \frac {2}{3\sqrt {3}} , -\frac {1}{\sqrt {3}}\right )$ .