3.6 Calculating Higher-Order Derivatives

3.6 Calculating Higher-Order Derivatives

A derivative is itself a function, so it can be differentiated again. The result, the second derivative, measures how a rate of change is itself changing: if position has velocity as its first derivative, then acceleration is its second. This page shows how to calculate these repeated derivatives with the rules selected in Topic 3.5, and how to write them correctly.

Higher-order derivatives

A higher-order derivative is found by differentiating a derivative again. If $y=f(x)$, then the first derivative is $y'$ or $\dfrac{dy}{dx}$, and the second derivative is

$$ y''=\frac{d^2y}{dx^2}. $$

More generally, the $n$th derivative is written as

$$ f^{(n)}(x) \qquad\text{or}\qquad \frac{d^ny}{dx^n}. $$

For

$$ y=x^4-3x^2+2x, $$

the first derivative is

$$ y'=4x^3-6x+2, $$

and differentiating again gives

$$ y''=12x^2-6. $$

A third derivative is

$$ y'''=24x, $$

and the fourth derivative is

$$ y^{(4)}=24. $$

The notation communicates what has happened: $\dfrac{d^2y}{dx^2}$ means “differentiate twice with respect to $x$,” not $\left(\dfrac{dy}{dx}\right)^2$. In applications, $y'$ can represent a rate of change, while $y''$ describes how that rate itself changes. Repeated differentiation uses the same rules as the first derivative.

A second derivative that needs more than the power rule. Topic 3.5 found that $y=(x^2+1)^3$ has $y'=6x(x^2+1)^2$. That derivative is a product whose second factor is a composition, so differentiating again takes the product rule and the chain rule together: $y''=6(x^2+1)^2+6x\cdot 2(x^2+1)(2x)=6(x^2+1)(5x^2+1)$.

Retrieval check

For $y=\ln(5x^2+1)$, identify the procedure and calculate $\dfrac{dy}{dx}$. Then state why a factor of $10x$ must appear. Finally, if $y'=10x/(5x^2+1)$, write $y''$ using $y''$ notation. The essential habit is: classify first, differentiate layer by layer, and check both the mathematics and the notation.