Introduction

Topic 3.6 is the final topic of AP Calculus AB Unit 3. After this topic you will have all the derivative knowledge needed in order to move on to Applications of Differentiation (Units 4 and 5). In this topic we will learn how to calculate higher-order derivatives, meaning the derivatives of derivatives.

How to Calculate Higher-Order Derivatives

The process used to calculate higher order derivatives is quite simple. All you have to do is literally take the derivative of a function and then take the derivative of what function which that differentiation yields. Literally take the derivative of the derivative. 

Find the second-order derivative of $f(x)=12x^3$.

The term “second-order” means to take the derivative twice. Normally, we would do first-order derivatives but this time we will have to differentiate it twice. In general, the $n$th order derivative is the result of differentiating a function $n$ times.

The derivative is $f^{\prime }(x)=36x^2$, and so we take the derivative of this again to get $f^{\prime \prime }(x)=72x$. Notice how we add another apostrophe to show that we took the derivative again.

The process used to calculate higher order derivatives is quite simple. All you have to do is literally take the derivative of a function and then take the derivative of what function which that differentiation yields. I would like to illustrate this through an example, take the function:

$f(x)=12x^3$

Though this idea seems quite simple and its applications might not seem immediately apparent, this concept of higher order-derivatives will come in handy later on in Calculus AB. For example, taking the $(n+1)$th order derivative of a $n$th-degree polynomial function will render that function equal to zero.

Notation Used for Higher-Order Derivatives

Like regular derivatives, there are many ways of notating higher-order derivatives. Lagrange’s notation is the type of notation I used in the example calculation, where a function is denoted as $f$, its derivative as $f^{\prime }$, second derivative as $f^{\prime \prime }$, third as $f^{\prime \prime \prime }$ and so on. These are read as “f prime”, “f double prime”, and “f triple prime” respectively. For any $n$th order derivative it can be written as $f^{(n)}(x)$ in Lagrange's notation. A variation of Lagrange’s notation you may see is where $y$ is your function, $y^{\prime }$ is your derivative and so on. You may find that this notation becomes impractical when you have a derivative with an order higher than around three, for the 8th order derivative you would end up with something looking like this: $f^{\prime \prime \prime \prime \prime \prime \prime \prime }(x)$.

Another notation for higher-order derivatives you will see is Leibniz’s notation. Leibniz’s for derivative works as follows: Where $y$ is your function, $\frac{dy}{dx}$ is its derivative, its second derivative is $\frac{d^{2}y}{d{x}^2}$, and so on again. This can be generalized as $\frac{d^{n}y}{d{x}^n}$ for any $n$th-order derivative. What makes this notation so useful is how it shows what variable you are differentiating with respect to (i.e. the $dx$ in the denominator). This way of expressing a derivative as a “fraction” will not only come in handy in further math classes but it will come in handy later on in this class.

There is a final type of notation you may see used for higher-order derivatives but this one is a bit rarer, that being Newton’s notation. It should also be noted that you will not see this in AP Calculus AB or BC. Using Newton’s notation $y$ represents your function, $\dot{y}$ is your derivative and $\ddot{y}$ is your second derivative. As I said before, this notation is quite rare and you most likely will not see it in Calculus AB. It is useful to know though since it may show up in later math or physics classes you take.

That might seem like a lot but condensed down into table it’s quite a bit clearer:

$\begin{array}{ccc}& \text{First Order Derivative}& n\text{th Order Derivative}\\ \text{Lagrangian Notation}& f^{\prime }(x)& f^{(n)}(x)\\ \text{Leibniz Notation}& \frac{dy}{dx}& \frac{d^{n}y}{d{x}^n}\\ \text{Newtonian Notation}& \dot{y}& \stackrel{(n)}{y}\end{array}$

More on Calculating Higher-Order Derivatives

I would like to go over a few more examples of calculating higher order derivatives before providing practice problems. Let’s take a function that is a bit harder to get the second derivative of, for example: $y=e^{2x^3}\tan (3x)$

We take the first derivative through differentiation:

$\frac{dy}{dx}=\frac{d}{dx}[e^{2x^3}\tan (3x)]$

$\frac{dy}{dx}=(e^{2x^3}\cdot 6x^2)\tan (3x)+e^{2x^3}(3{\sec }^2(3x))$

$\frac{dy}{dx}=6x^2{e}^{2x^3}\tan (3x)+e^{2x^3}3{\sec }^2(3x)$

$\frac{dy}{dx}=e^{2x^3}(6x^2\tan (3x)+3{\sec }^2(3x))$

Now we take the derivative again to get the second-order derivative.

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}[{\sec }^2(3x)(6x^2{e}^{2x^3}+3e^{2x^3})]$

$\frac{d^{2}y}{d{x}^2}=6x^2{e}^{2x^3}(6x^2\tan (3x)+3{\sec }^2(3x))+e^{2x^3}(18{\sec }^2(3x)\tan (3x)+12x\tan (3x)+18x^2{\sec }^2(3x))$

This could be potentially further simplified but that is our answer for the second derivative. This function, unlike our first example, became much harder to differentiate as we took the second derivative. At this point to take the third derivative may take upwards of 10 or 20 minutes with how many terms we have. So let's try again with another function, but this time take the third-order derivative. We will use the function $g(x)=3x^2+15x^2-21x$. This function will be relatively simple to differentiate using the power rule since it’s a polynomial

$g(x)=3x^3+15x^2-21x$

Take the first derivative by applying the power rule:

$g^{\prime }(x)=9x^2+30x-21$

Take the second derivative by applying the power rule:

$g^{\prime \prime }(x)=18x+30$

Finally take the third derivative by applying the power rule: $g^{\prime \prime \prime }(x)=18$

Our final answer for the third-order derivative of $g(x)=3x^3+15x^2-21x$ comes out to be $g^{\prime \prime \prime }(x)=18$. You will more likely see polynomial functions like this or functions that differentiate cyclically when you are asked to calculate higher-order derivatives. It is still useful to know how to take higher-order derivatives of all types of functions though.

Let's do one more problem to solidify what we’ve learnt.  

Find $g^{\prime \prime \prime }(\pi )$ if $g(x)=x^4+0.5\sin (x)$. 

Take the first derivative to get $g^{\prime }(x)=4x^3+0.5\cos (x)$.

Take the second derivative to get $g^{\prime \prime }(x)=12x^2-0.5\sin (x)$.

Finally take the third derivative to get $g^{\prime \prime \prime }(x)=24x-0.5\cos (x)$.

We can now plug in $x=\pi$ and get the answer $g’’’(x)=24\pi -0.5\cos (\pi )=24\pi$

Implicit Differentiation

For implicit differentiation, it is what you would expect. 

Solve for $\frac{d^{2}y}{d{x}^2}$ if $x{y}^3=x^2-y^2$.

To solve this, we can go with the standard procedure of implicit differentiation.

$x{y}^3=x^2-y^2$

$y^3+3x{y}^2\cdot \frac{dy}{dx}=2x-2y\cdot \frac{dy}{dx}$

$y^3-2x=-2y\cdot \frac{dy}{dx}-3x{y}^2\cdot \frac{dy}{dx}$

$\frac{y^3-2x}{-2y-3xy}=\frac{dy}{dx}$

We have the first derivative, and we can differentiate again for the second derivative.

$\frac{y^3-2x}{-2y-3xy}=\frac{dy}{dx}$

$\frac{(3y^2\cdot \frac{dy}{dx}-2)(-2y-3xy)-(-2y’-3y-3x\cdot \frac{dy}{dx})(y^3-2x)}{(-2y-3xy{)}^2}=\frac{d^{2}y}{d{x}^2}$

Notice that in our second-order derivative, we have $\frac{dy}{dx}$. However, this is not acceptable in a derivative. You cannot have a derivative in a derivative, so we will substitute in the expression we found for $\frac{dy}{dx}$ into the equation.

$\frac{d^{2}y}{d{x}^2}=\frac{(3y^2\cdot \frac{y^3-2x}{-2y-3xy}-2)(-2y-3xy)-(-2\frac{y^3-2x}{-2y-3xy}-3y-3x\cdot \frac{y^3-2x}{-2y-3xy})(y^3-2x)}{(-2y-3xy{)}^2}$

Given $x^2+y^2=36$, find $\frac{d^{2}y}{d{x}^2}$:

$2x+2y\cdot \frac{dy}{dx}=0$

$2y\cdot \frac{dy}{dx}=-2x$

$\frac{dy}{dx}=\frac{-2x}{2y}⟹-\frac{x}{y}$

Now, we will differentiate again.

$\frac{d^{2}y}{d{x}^2}=-\left(\frac{y-x\cdot \frac{dy}{dx}}{y^2}\right)$

$\frac{d^{2}y}{d{x}^2}=-\left(\frac{y-x\cdot -\frac{x}{y}}{y^2}\right)$

$\frac{d^{2}y}{d{x}^2}=-\left(\frac{y+\frac{x^2}{y}}{y^2}\right)$

$\frac{d^{2}y}{d{x}^2}=-\left(\frac{\frac{y^2+x^2}{y}}{y^2}\right)$

$\frac{d^{2}y}{d{x}^2}=-\left(\frac{y^2+x^2}{y^3}\right)$

Since $x^2+y^2=36$, please replace! In general, if you notice that there are components in your derivative that resembles other expressions that were already defined, do rewrite them for the sake of simplicity! Be sure to check!

$\frac{d^{2}y}{d{x}^2}=-\frac{36}{y^3}$

Notes on Higher-Order Derivatives

Practice Questions