Second and higher-order derivatives

The derivative $f'$ is a function in its own right: it takes an input $x$ and returns the rate of change of $f$ there. Like any function, it can be differentiated. Doing so gives a new function, one that measures how fast the rate of change is itself changing rather than how fast $f$ is changing.

The symbol $\dfrac{d^2y}{dx^2}$ records that the operator $\dfrac{d}{dx}$ has been applied twice. It is a single piece of notation rather than a power, so it does not mean $\dfrac{dy}{dx}$ squared.

Computing $f''$ needs no new rules. The same differentiation rules apply to $f'$ as applied to $f$.

Which rules apply when differentiating $f'$ is decided by the form of $f'$, not by the form of $f$. Differentiating can change that form, so we read the structure of $f'$ before differentiating it, exactly as we would for any other function.

Compute $f''(x)$ for $f(x) = x^2e^x$.

Solution

1. We differentiate $f$ to obtain $f'$. The expression is a product of two factors, so we name them:

   $$u(x) = x^2, \qquad v(x) = e^x.$$

   Each factor differentiates on its own: the power rule gives $u'(x) = 2x$, and $v'(x) = e^x$. Assembling with the product rule,

   $$\begin{align*} f'(x) &= u'(x)\,v(x) + u(x)\,v'(x) \\ &= 2xe^x + x^2e^x. \end{align*}$$

2. We simplify $f'$. Both terms carry a factor of $e^x$, so we factor it out:

   $$f'(x) = \left(x^2 + 2x\right)e^x.$$

   Written this way, the structure of $f'$ is visible. It is again a product, but not the product we began with: its first factor is $x^2 + 2x$, where the first factor of $f$ was $x^2$.

3. We differentiate $f'$ to obtain $f''$. Its structure requires the product rule again, now with the factors

   $$p(x) = x^2 + 2x, \qquad q(x) = e^x,$$

   giving $p'(x) = 2x + 2$ and $q'(x) = e^x$. Assembling with the product rule again,

   $$\begin{align*} f''(x) &= p'(x)\,q(x) + p(x)\,q'(x) \\ &= (2x + 2)e^x + \left(x^2 + 2x\right)e^x \\ &= \left(x^2 + 4x + 2\right)e^x. \end{align*}$$

The second derivative is not the end of the process. Differentiating $f''$ gives the third derivative $f'''(x)$, that function can be differentiated in turn to give a fourth, and so on for as far as we choose to go.

For a polynomial, we can find the order at which the derivatives vanish, and the degree of every derivative below it, without computing them. The power rule lowers each exponent by one, so the highest power falls by one at every differentiation. Once the degree reaches $0$ the derivative is a non-zero constant, and differentiating a constant gives $0$, so every derivative beyond that order is $0$.

Only the leading term of the polynomial matters. Lower-degree terms reach $0$ at earlier orders of differentiation, so they are never the last to vanish.

Find $f^{(4)}(x)$ for $f(x) = x^5 - 2x^3 + 7x$, and state the smallest order at which the derivatives become $0$.

Solution

We differentiate four times, applying the power rule to each term:

So $f^{(4)}(x) = 120x$, and the degree falls by one at each order, as the power rule requires.

The order at which the derivatives vanish follows from the degree alone. The polynomial has degree $5$, so the derivative of order $5$ has degree $0$:

a non-zero constant. Differentiating it gives $f^{(6)}(x) = 0$, and every derivative beyond the sixth is $0$ as well. The smallest such order is $6$, which the degree tells us directly, without computing the table above.