Second Derivative Test

The second derivative test is used to determine if a given stationary point is a maximum or minimum.

The first step of the second derivative test is to find stationary points.

A stationary point is a point on a curve with a gradient of zero such that \(f'(x) = 0\) or \( \frac{dy}{dx} = 0\). \( _\square \)

Find the stationary points of the curve \(y = \frac{1}{3}x^3 + 4x^2 - 20x\).

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We have \( \frac{dy}{dx} = x^2 + 8x - 20.\) Since stationary points occur when \( \frac{dy}{dx} = 0\), the values of \(x\) for which \( \frac{dy}{dx} = 0\) are obtained as follows:

\[\begin{align} x^2 + 8x - 20 &= 0\\ (x + 10)(x - 2) &= 0\\ x &= -10,~ x = 2. \end{align}\]

Substituting these values of \(x\) back into the original curve equation gives the full pairs of coordinates:

\[\begin{align} y &= \frac{1}{3}(-10)^3 + 4(-10)^2 - 20(-10)\\& = -\frac{1000}{3} + 400 + 200 \\&= \frac{800}{3}\quad \\\\ y&= \frac{1}{3}(2)^3 + 4(2)^2 - 20(2) \\&= \frac{8}{3} + 16 - 40 \\&= -\frac{64}{3}, \end{align}\]

which implies that the stationary points are \(\left(-10, \frac{800}{3}\right)\) and \(\left(2, -\frac{64}{3}\right).\) \( _\square \)

Note in the example above that the full coordinates were found. When dealing with the second derivative test, only the \(x\)-coordinate values are needed.

A second derivative is found by differentiating a derivative of a function. In other words, differentiate a function to get its derivative, then differentiate again to get its second derivative.

Notation for second derivatives can be represented as

• \(f''(x)\), read as "f double-dashed x";
• \( \frac{d^2y}{dx^2}\), read as "d two y by d x squared."

Find the second derivative of \(f(x) = 3x^4 + x^3 + 2x^2 + 17\).

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Differentiate the first time to get the derivative: \(f'(x) = 12x^3 + 3x^2 + 4x.\)

Now, differentiate the derivative to get the second derivative: \(f''(x) = 36x^2 + 6x + 4\). \( _\square \)

Given that \(p = -4q^5 - 3q^3 + q^2\), find \( \frac{d^2p}{dq^2}\).

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First, find the derivative of \(p:\) \( \frac{dp}{dq} = -20q^4 - 9q^2 + 2q\).

Now, differentiate \( \frac{dp}{dq}\) to find the second derivative: \( \frac{d^2p}{dq^2} = -80q^3 - 18q + 2\). \( _\square \)

After finding the \(x\)-values of the stationary points, the second derivative needs to be found. Then, we substitute the \(x\)-values from the stationary points into the second derivative to perform the test. The test has three outcomes:

• If the second derivative is less than zero, the stationary point is a maximum.
• If the second derivative is greater than zero, the stationary point is a minimum.
• If the second derivative equals zero, the stationary point could be a point of inflection.

The derivative of a curve is found to be \(g'(x) = 4x^2 - \frac{7}{2}x\). Stationary points on this curve occur when \(x = 0\) and \(x = \frac{7}{8}\). Determine if these points are maxima or minima using the second derivative test.

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We have \(g''(x) = 8x - \frac{7}{2}\).
Now, substitute \(x = 0\) to determine if it is a maximum or minimum point: \(g''(0) = 8(0) - \frac{7}{2} = -\frac{7}{2}\).
As \(g''(0) < 0\), the stationary point when \(x = 0\) is a maximum.

Do the same for the other value: \(g''\left(\frac{7}{8}\right) = 8\left(\frac{7}{8}\right) - \frac{7}{2} = 7 - \frac{7}{2} = \frac{7}{2}\).
As \(g''\left(\frac{7}{8}\right) > 0\), the stationary point when \(x = \frac{7}{8}\) is a minimum. \( _\square \)

• Differentiation Rules
• Implicit Differentiation
• Higher-order Derivatives