Implicit differentiation
Tonight's idea
The circle \(x^2 + y^2 = 25\) is not a function: most \(x\) give two \(y\). But near any point except \((\pm 5, 0)\) it behaves like one; on the upper half, \(y = \sqrt{25 - x^2}\). Rather than solve for \(y\), keep it implicit. Treat \(y\) as a function of \(x\) and differentiate both sides. The chain rule turns \(y^2\) into \(2y \cdot y'\), where \(y'\) means \(dy/dx\):
\(2x + 2y\,y' = 0\), so \(y' = -\dfrac{x}{y}\).
Geometry agrees. The radius to \((x, y)\) has slope \(y/x\); the tangent is perpendicular, so its slope is \(-x/y\). At \((3, 4)\): slope \(-\tfrac34\), tangent \(3x + 4y = 25\).
For any equation in \(x\) and \(y\): differentiate every term, writing \(y'\) whenever you differentiate something containing \(y\), then solve for \(y'\). The answer usually involves both \(x\) and \(y\), so the point matters.
The same move handles \(x^x\), neither a power (the exponent varies) nor an exponential (the base varies). Take logs: \(\ln y = x \ln x\). Differentiate, chain rule on the left, product rule on the right: \(y'/y = \ln x + 1\). So
\(\dfrac{d}{dx}\, x^x = x^x(\ln x + 1)\), zero at \(x = 1/e\), where \(x^x\) is smallest.
Watch one: the slope of \(x^3 + y^3 = 6xy\) at \((3, 3)\)
First check the point is on the curve: \(27 + 27 = 54 = 6 \cdot 9\). Differentiate every term (the right side needs the product rule), then collect the \(y'\) terms:
\[3x^2 + 3y^2\,y' = 6y + 6x\,y' \quad\Longrightarrow\quad y' = \frac{2y - x^2}{y^2 - 2x} = \frac{6 - 9}{9 - 6} = -1 \ \text{ at } (3, 3).\]
The tangent is \(x + y = 6\). Nobody could have started by solving for \(y\): this curve, the folium of Descartes, has no usable formula \(y = f(x)\). Implicit differentiation never asked for one.
Your turn
- For \(xy = 4\), find \(y'\) implicitly. Then solve for \(y\), differentiate directly, and check the two answers agree.
- Find the slope of \(x^2 - xy + y^2 = 7\) at \((1, 3)\), and the equation of the tangent there.
- (a) Differentiate \(y = x^{1/x}\) by taking logs first. (b) Find where \(y' = 0\). Hence decide which of \(\sqrt2\), \(\sqrt[3]{3}\) and \(e^{1/e}\) is largest, without a calculator.
- For the circle \(x^2 + y^2 = 25\), differentiate \(y' = -x/y\) once more (quotient rule, remembering \(y'\) inside) and use the circle's equation to show \(y'' = -25/y^3\). What does the sign of \(y''\) say about the upper half and the lower half?
Stretch: where the folium is flat
Where is the tangent to \(x^3 + y^3 = 6xy\) horizontal? Set the numerator of \(y'\) to zero, substitute into the curve, and solve. Give exact coordinates. What goes wrong at the origin?
Answers
- \(y + x\,y' = 0\), so \(y' = -y/x\). Directly, \(y = 4/x\) gives \(y' = -4/x^2\), and since \(y = 4/x\) this is \(-y/x\). Same thing.
- \(2x - y - x\,y' + 2y\,y' = 0\), so \(y' = \dfrac{y - 2x}{2y - x} = \dfrac{3 - 2}{6 - 1} = \tfrac15\). Tangent: \(y - 3 = \tfrac15(x - 1)\), or \(x - 5y + 14 = 0\).
- (a) \(\ln y = \dfrac{\ln x}{x}\), so \(\dfrac{y'}{y} = \dfrac{1 - \ln x}{x^2}\) and \(y' = x^{1/x}\,\dfrac{1 - \ln x}{x^2}\). (b) \(y' = 0\) when \(\ln x = 1\), at \(x = e\); the derivative is positive before and negative after, so \(x = e\) is the maximum of \(x^{1/x}\). Hence \(e^{1/e}\) is the largest of the three (about 1.4447, against \(\sqrt2 \approx 1.4142\) and \(\sqrt[3]{3} \approx 1.4422\)).
- \(y'' = -\dfrac{y - x\,y'}{y^2} = -\dfrac{y + x^2/y}{y^2} = -\dfrac{y^2 + x^2}{y^3} = -\dfrac{25}{y^3}\). On the upper half \(y > 0\), so \(y'' \lt 0\): concave down, a cap. On the lower half \(y'' > 0\): concave up, a bowl.
Stretch. \(2y = x^2\), so \(y = x^2/2\). Substituting: \(x^3 + x^6/8 = 3x^3\), so \(x^6 = 16x^3\), giving \(x = 0\) or \(x^3 = 16\). The horizontal tangent is at \(x = \sqrt[3]{16}\), \(y = x^2/2 = \sqrt[3]{32}\), about \((2.52, 3.17)\). At the origin both numerator and denominator of \(y'\) vanish: the formula gives \(0/0\). The curve crosses itself there and has two tangents, the axes, so no single slope exists.