Find an implicit second derivative
How the slope of an ellipse changes
The curve \(x^2 + 4y^2 = 20\) is an ellipse. The orange line is the tangent, and its slope is \(\tfrac{dy}{dx} = -\tfrac{x}{4y}\).
Drag the point across the top half, from left to right. The slope falls from positive to negative, and the rate at which the slope changes is the second derivative \(y''\).
The strip between the curve and its tangent is green where \(y'' > 0\) and red where \(y'' < 0\). On the top half the slope falls, so \(y'' < 0\) there and the strip is red.
To find \(y''\), you differentiate \(y'\) again.
Differentiate y′ again
Drag the point along the curve. The slope \(\cy{y'}\) of the tangent changes, and the second derivative \(y''\) is the rate at which it changes.
The formula for \(\cy{y'}\) contains \(x\) and \(y\), and both change when the point moves, because \(y\) depends on \(x\). The derivative of \(\cy{y'}\) therefore needs the chain rule again, and the result contains \(\cy{y'}\).
Notation
Differentiate both sides to find \(\cy{y'}\), which means \(\tfrac{dy}{dx}\). The chain rule turns \(4y^2\) into \(8y\,\cy{y'}\), so \(2x + 8y\,\cy{y'} = 0\).
\[x^2 + 4y^2 = 20 \;\Rightarrow\; \cy{y'} = -\frac{x}{4y}\]Use the quotient rule to differentiate \(\cy{y'}\) with respect to \(x\). Because \(y\) depends on \(x\), the derivative of \(y\) is \(\cy{y'}\).
\[y'' = -\frac{4y - 4x\,\cy{y'}}{16y^2} = -\frac{y - x\,\cy{y'}}{4y^2}\]Replace \(\cy{y'}\) by \(-\tfrac{x}{4y}\). Then multiply the top and the bottom by \(4y\) to clear the fraction inside the fraction.
\[\begin{aligned} y'' &= -\frac{y + \frac{x^2}{4y}}{4y^2} \\ &= -\frac{x^2 + 4y^2}{16y^3} \end{aligned}\]On the curve, \(x^2 + 4y^2 = 20\), so replace \(x^2 + 4y^2\) by 20. The result holds only at points of the curve where \(y \neq 0\).
\[y'' = -\frac{20}{16y^3} = -\frac{5}{4y^3}\]
When it fails
The last step used the equation of the curve, so \(-\tfrac{5}{4y^3}\) is correct only at points of the curve. Put a point off the curve, at \((2, 1)\).
At \((2, 1)\), \(x^2 + 4y^2 = 8\), not 20, so the two forms give different numbers: \(-\tfrac{x^2 + 4y^2}{16y^3} = -\tfrac12\), but \(-\tfrac{5}{4y^3} = -\tfrac54\).
Example 1: A parabola on its side
Find \(\tfrac{d^2y}{dx^2}\) on the curve \(y^2 = 4x\).
Differentiate both sides to find \(\cy{y'}\). The chain rule turns \(y^2\) into \(2y\,\cy{y'}\).
\[2y\,\cy{y'} = 4 \;\Rightarrow\; \cy{y'} = \frac{2}{y}\]Differentiate \(\cy{y'} = \tfrac{2}{y}\). The chain rule gives a factor \(\cy{y'}\), because \(y\) depends on \(x\).
\[y'' = -\frac{2}{y^2}\,\cy{y'}\]Replace \(\cy{y'}\) by \(\tfrac{2}{y}\).
\[y'' = -\frac{4}{y^3}\]As a check, the top half is \(y = 2\sqrt{x}\), and its second derivative is \(-\tfrac12x^{-3/2}\), which is \(-\tfrac12\) at \(x = 1\). The formula gives the same value at \((1, 2)\), where the red strip shows \(y'' < 0\).
\[-\frac{4}{2^3} = -\frac12\]
Example 2: A hyperbola
Find \(y''\) on the curve \(x^2 - y^2 = 7\), and its value at \((4, 3)\).
Differentiate both sides to find \(\cy{y'}\).
\[2x - 2y\,\cy{y'} = 0 \;\Rightarrow\; \cy{y'} = \frac{x}{y}\]A tempting wrong move: Differentiate \(\tfrac{x}{y}\) and treat \(y\) as a constant.
\[y'' = \frac{1}{y} = \frac13\]Holding \(y\) constant moves the point along the red horizontal line, off the curve. The value \(\tfrac13\) is positive, which would make the strip green, but the strip at \((4, 3)\) is red.
Differentiate \(\tfrac{x}{y}\) with the quotient rule. Because \(y\) depends on \(x\), the derivative of \(y\) is \(\cy{y'}\).
\[y'' = \frac{y - x\,\cy{y'}}{y^2}\]Replace \(\cy{y'}\) by \(\tfrac{x}{y}\), and multiply the top and the bottom by \(y\). On the curve \(x^2 - y^2 = 7\), so replace \(y^2 - x^2\) by \(-7\).
\[\begin{aligned} y'' &= \frac{y - \frac{x^2}{y}}{y^2} = \frac{y^2 - x^2}{y^3} \\ &= -\frac{7}{y^3} \end{aligned}\]Substitute \(y = 3\). The value is negative, so the curve lies below its tangent at \((4, 3)\), and the strip there is red.
\[y''(4, 3) = -\frac{7}{27}\]
Practice
Find \(y''\) on the curve \(x^3 + y^3 = 9\). Write the answer with \(x\) and \(y\) only.
Find \(y'\) first. Use the quotient rule on it, and replace \(y'\). At the end, replace \(x^3 + y^3\) by 9.
Solution
Differentiate both sides to find \(\cy{y'}\).
\[3x^2 + 3y^2\,\cy{y'} = 0 \;\Rightarrow\; \cy{y'} = -\frac{x^2}{y^2}\]Differentiate \(\cy{y'}\) with the quotient rule. Because \(y\) depends on \(x\), the derivative of \(y^2\) is \(2y\,\cy{y'}\).
\[y'' = -\frac{2xy^2 - x^2\cdot2y\,\cy{y'}}{y^4}\]Replace \(\cy{y'}\) by \(-\tfrac{x^2}{y^2}\), and multiply the top and the bottom by \(y\).
\[\begin{aligned} y'' &= -\frac{2xy^3 + 2x^4}{y^5} \\ &= -\frac{2x\,(y^3 + x^3)}{y^5} \end{aligned}\]On the curve, \(x^3 + y^3 = 9\), so replace \(x^3 + y^3\) by 9.
\[y'' = -\frac{18x}{y^5}\]
Questions
The graph shows the ellipse \(x^2 + 4y^2 = 20\), where \(y'' = -\tfrac{5}{4y^3}\). The orange point is \(A = (2, -2)\), with its tangent and the strip between the curve and the tangent. The tangent at \(B = (4, 1)\) is also drawn.
What is \(y''\) at the point \(A = (2, -2)\)? The strip at \(A\) is green.
- \(\tfrac{5}{32}\)
- \(-\tfrac{5}{32}\)
- \(\tfrac58\)
Answer
\(\tfrac{5}{32}\)
At \(B = (4, 1)\) the orange tangent has slope \(-1\). What value does \(y'' = -\tfrac{y - x\,y'}{4y^2}\) give there?
- \(-\tfrac54\)
- \(\tfrac54\)
- \(\tfrac34\)
Answer
\(-\tfrac54\)
Find \(y''\) for the curve \(x^2 + xy = 4\). Write the answer with \(x\) and \(y\) only.
- \(\dfrac{2(x + y)}{x^2}\)
- \(-\dfrac{2(x + y)}{x^2}\)
- \(\dfrac{y}{x^2}\)
- \(\dfrac{2(x + y)}{x}\)
Answer
\(\dfrac{2(x + y)}{x^2}\)
Summary
You find \(y'\), differentiate it again with the chain rule, replace \(y'\), and simplify with the equation of the curve.
This works because \(y\) depends on \(x\), so the derivative of the formula for \(y'\) contains \(y'\) again.
You use it to find how a curve bends when the curve is given by an equation in \(x\) and \(y\).
This is one of four skills in topic 5.12 of Which & Why: AP Calculus.