Find an implicit first derivative
The slope of a circle
The circle \(x^2 + y^2 = 17\) is not the graph of one function of \(x\), because the dashed vertical line meets it twice. The top half is one function and the bottom half is another.
Drag the orange point around the circle. The orange line is the tangent, and its slope is shown under the figure.
The readout \(-\tfrac{x}{y}\) matches the slope at every point. At \((4, 1)\) both are \(-4\), so the slope depends on both coordinates of the point.
You can find this slope without solving the equation for \(y\).
y depends on x
Near the orange point, the curve is the graph of a function \(y\) of \(x\), even though the whole curve is not. The toggle shows that piece as a thick line.
The slope \(\tfrac{dy}{dx}\) of the tangent is the slope of that piece. When \(x\) changes, \(y\) changes too, so the derivative of an expression in \(y\) needs the chain rule, which brings a factor \(\tfrac{dy}{dx}\).
This method is called implicit differentiation, because the equation determines \(y\) without a formula for it.
Notation
The slope of the tangent is the rise \(\cy{dy}\) over the run \(dx\). Both sides are equal at every point of the curve, so their derivatives with respect to \(x\) are equal too.
\[\frac{d}{dx}\left(x^2 + y^2\right) = \frac{d}{dx}(17)\]The term \(x^2\) gives \(2x\). Because \(y\) depends on \(x\), the chain rule turns \(y^2\) into \(2y\,\cy{\tfrac{dy}{dx}}\). The constant 17 gives 0.
\[2x + 2y\,\cy{\frac{dy}{dx}} = 0\]A product such as \(xy\) needs the product rule. Both of its factors change with \(x\), so it also brings \(\cy{\tfrac{dy}{dx}}\).
\[\frac{d}{dx}(xy) = y + x\,\cy{\frac{dy}{dx}}\]To solve for \(\cy{\tfrac{dy}{dx}}\), move \(2x\) to the right side, so \(2y\,\cy{\tfrac{dy}{dx}} = -2x\). Then divide by \(2y\).
\[\cy{\frac{dy}{dx}} = -\frac{x}{y}\]The point must lie on the curve, because the formula is true only there. Substitute its coordinates.
\[\left.\cy{\frac{dy}{dx}}\right|_{(4,\,1)} = -\frac{4}{1} = \cy{-4}\]
When it fails
The formula needs \(y \neq 0\), because the last step divides by \(2y\). Move the point to \((\sqrt{17}, 0)\).
There \(y = 0\), so \(-\tfrac{x}{y}\) has a zero denominator. The tangent is vertical, and a vertical line has no slope. The dashed line \(x = 4\) meets the circle twice, so near this point \(y\) is not a function of \(x\).
Example 1: A hyperbola
Find \(\tfrac{dy}{dx}\) on the curve \(xy = 6\), and the slope at \((2, 3)\).
Differentiate both sides with respect to \(x\). The left side is a product, so use the product rule. The constant 6 gives 0.
\[y + x\,\cy{\frac{dy}{dx}} = 0\]To solve for \(\cy{\tfrac{dy}{dx}}\), subtract \(y\) from both sides. Then divide by \(x\).
\[\cy{\frac{dy}{dx}} = -\frac{y}{x}\]The point \((2, 3)\) is on the curve, because \(2\cdot3 = 6\). Substitute. In the figure, a chalk run of 1 to the left meets an orange rise of \(\tfrac32\), so the slope is \(-\tfrac32\).
\[\cy{\frac{dy}{dx}} = -\frac{3}{2}\]
Example 2: A product and a square
Find the slope of the curve \(x^2y + y^2 = 6\) at \((1, 2)\).
The point \((1, 2)\) is on the curve, because \(1^2\cdot2 + 2^2 = 6\). Differentiate both sides. The term \(x^2y\) needs the product rule, and \(y^2\) needs the chain rule. Below, \(y'\) stands for \(\tfrac{dy}{dx}\).
A tempting wrong move: The derivative of \(y^2\) is \(2y\).
\[\begin{aligned} &2xy + x^2\,y' + 2y = 0 \\ &y' = -\frac{2xy + 2y}{x^2} = -8 \end{aligned}\]Because \(y\) depends on \(x\), the chain rule requires a factor \(\tfrac{dy}{dx}\), and this leaves it out. The red line with slope \(-8\) crosses the curve at \((1, 2)\) instead of touching it, so it is not the tangent.
The derivative of \(y^2\) is \(2y\,\cy{y'}\). Factor \(\cy{y'}\) out of its two terms. Then divide by \(x^2 + 2y\).
\[\begin{aligned} 2xy + x^2\,\cy{y'} + 2y\,\cy{y'} &= 0 \\ \cy{y'} &= -\frac{2xy}{x^2 + 2y} \end{aligned}\]Substitute \(x = 1\) and \(y = 2\). The orange line with this slope touches the curve at \((1, 2)\), so it is the tangent.
\[\cy{\frac{dy}{dx}} = -\frac{2\cdot1\cdot2}{1 + 4} = \cy{-\tfrac45}\]
Practice
Find \(\tfrac{dy}{dx}\) on the curve \(xe^y + y = 1\), and the slope at \((1, 0)\).
Use the product rule on \(xe^y\). The derivative of \(e^y\) is \(e^y\) times \(\tfrac{dy}{dx}\). Then drag the point to \((1, 0)\) and compare the slope with your answer.
Solution
Differentiate both sides. The term \(xe^y\) needs the product rule, and \(e^y\) needs the chain rule.
\[e^y + xe^y\,\cy{\frac{dy}{dx}} + \cy{\frac{dy}{dx}} = 0\]Factor \(\cy{\tfrac{dy}{dx}}\) out of its two terms. Then divide by \(xe^y + 1\).
\[\cy{\frac{dy}{dx}} = -\frac{e^y}{xe^y + 1}\]The point \((1, 0)\) is on the curve, because \(1\cdot e^0 + 0 = 1\). Substitute. In the figure, the orange rise is half the chalk run to the left, so the tangent has slope \(-\tfrac12\).
\[\cy{\frac{dy}{dx}} = -\frac{1}{1 + 1} = \cy{-\tfrac12}\]
Questions
Questions 1 and 2 use the curve \(xy^2 = 8\), which has an upper branch and a lower branch.
Which formula for \(\tfrac{dy}{dx}\) gives the slope at the point \(A = (2, 2)\) on the upper branch?
- \(-\dfrac{y}{2x}\)
- \(-\dfrac{y}{x}\)
- \(-\dfrac{y^2}{x}\)
- \(\dfrac{y}{2x}\)
Answer
\(-\dfrac{y}{2x}\)
What is the slope at the point \(B = (2, -2)\) on the lower branch?
- \(\tfrac12\)
- \(-\tfrac12\)
- \(1\)
- \(-1\)
Answer
\(\tfrac12\)
Find \(\tfrac{dy}{dx}\) for the curve \(x^3 + y^4 = 2xy\).
- \(\dfrac{2y - 3x^2}{4y^3 - 2x}\)
- \(\dfrac{2y - 3x^2}{4y^3}\)
- \(\dfrac{2y + 2x - 3x^2}{4y^3}\)
- \(\dfrac{3x^2 - 2y}{4y^3 - 2x}\)
Answer
\(\dfrac{2y - 3x^2}{4y^3 - 2x}\)
Summary
You differentiate both sides with respect to \(x\), treat \(y\) as a function of \(x\), and solve for \(\tfrac{dy}{dx}\).
This works because \(y\) depends on \(x\) near the point, so each term in \(y\) brings the factor \(\tfrac{dy}{dx}\).
You use it when the equation is hard to solve for y, or when it gives more than one function.
This is one of four skills in topic 5.12 of Which & Why: AP Calculus.