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6.1: Higher Order Derivatives

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    You may recall hearing about Becca and her Track and Field competition in a prior concept. Her boyfriend had been recording on her laptop the position signals from the GPS transmitter Becca was wearing during her race. She was using a laptop program she had written to process the once per second GPS position signals so she could determine how far she had run at time during the race. These distance and time points were plotted on her laptop. She had created a program that also gave her a mathematical model of her distance so that she could take the derivative of the math function at any time point to get her “instantaneous” speed.

    What if, instead of just finding her speed at any time during the race, she wanted to find her acceleration? How would that be done?

    Higher Order Derivatives

    If the function f has a derivative f′ that is differentiable, then the derivative of f′, denoted by f′′ is called the second derivative of f. We can continue the process of differentiating derivatives and obtain third, fourth, fifth and higher derivatives of f. They are denoted as shown below:

    Notations for Higher Order Derivatives
    1st 2nd 3rd 4th nth order
    f′ f′′ f′′′ f(4) f(n)
    y′ y′′ y′′′ y(4) y(n)
    \[ \frac{dy}{dx} \nonumber\] \[ \frac{d^2y}{dx^2} \nonumber\] \[ \frac{d^3y}{dx^3} \nonumber\] \[ \frac{d^4y}{dx^4} \nonumber\] \[ \frac{d^ny}{dx^n} \nonumber\]

    \[ D_xy \nonumber\]
    \[ D_x^2y\nonumber\] \[ D_x^3y \nonumber\] \[ D_x^4y\nonumber\] \[ D_x^ny \nonumber\]

    Given f(x)=−2x2−4x−1. What is f′′(x)?

    Recall that f′′(x) means “The second derivative of f(x)”, or “The derivative of the derivative of f(x)”. The function f(x) must be differentiated twice as follows:

    \[ f′(x)=\frac{d}{dx}(−2x^2−4x−1) \nonumber\]

    …Determine the 1st derivate

    \[ f′(x)=−4x−4 \nonumber\]

    \[ f′′(x)=\frac{d}{dx}(−4x−4) \nonumber\]

    .…Determine the 2nd derivate.

    \[ f′′(x)=−4 \nonumber\]

    Therefore, \[ f′′(x)=−4 \nonumber\]


    Example 1

    Earlier, you were asked about Becca would find her acceleration. Since Becca has already created a program to calculate her instantaneous speed at a given point on the track by finding the derivative of the mathematical model to her GPS position data, she could then take the derivative of that function, the second derivative, to find her instantaneous acceleration at the same point in the race.

    By finding her instantaneous speed and acceleration at different points in the race, she can learn a lot about her performance during the race, and hopefully target areas she needs to work on to improve her overall success.

    Example 2

    Given f(x)=(−x4−4x3−5x2+3). Find f′′(x) when x=3.

    Again, the function f(x) must be differentiated twice; then the 2nd derivative must be evaluated:

    \[ f′(x)=\frac{d}{dx}(−x^4−4x^3−5x^2+3) \nonumber\]

    …Determine the 1st derivate

    \[ =−4x^3−12x^2−10x \nonumber\]

    \[ f′′(x)=\frac{d}{dx}(−4x^3−12x^2−10x) \nonumber\]

    .…Determine the 2nd derivate

    \[ =−12x^2−24x−10 \nonumber\]

    \[ f′(3)=−12(3)2^−24(3)−10 \nonumber\]

    .…Evaluate the 2nd derivate

    \[ =−190 \nonumber\]

    Therefore, \[ f′′(3)=−190\nonumber\]

    Example 3

    Show that y=x3+3x+2 satisfies the differential equation y′′′+xy′′−2y′=0.

    We need to obtain the first, second, and third derivatives and substitute them into the differential equation to check for equality.

    \[ y=x^3+3x+2 \nonumber\]

    \[ y′=3x^2+3 \nonumber\]

    \[ y′′=6x \nonumber\]

    \[ y′′′=6 \nonumber\]


    \[ y′′′+xy′′−2y′=6+x(6x)−2(3x^2+3) \nonumber\]

    \[ =6+6x^2−6x^2−6 \nonumber\]

    \[ =0 \nonumber\]

    which satisfies the equation.

    Example 4

    Find the fifth derivative of \[ f(x)=2x^4−3x^3+5x^2−x−1 \nonumber\]

    To find the fifth derivative, we must first find the first, second, third, and fourth derivatives as follows:

    1. \[ f′(x)=8x^3−9x^2+5x−x \nonumber\]
    2. \[ f(2)(x)=24x^2−18x+5 \nonumber\]
    3. \[ f(3)(x)=48x−18 \nonumber\]
    4. \[ f(4)(x)=48 \nonumber\]
    5. \[ f(5)(x)=0 \nonumber\]


    1. Given: \[ v(x)=−4x^3+3x^2+2x+3 \nonumber\] What is v′′(x)?
    2. Given: \[ m(x)=x^2+5x \nonumber\] What is m′′(x)?
    3. Given: \[ d(x)=3x^4e^x \nonumber\] What is d′′(x)?
    4. Given: \[ t(x)=−2x^5sin(x) \nonumber\] What is \[ \frac{d^2t}{dx^2}? \nonumber\]
    5. What is \[ \frac{d^2}{dx^2}3x^5e^x? \nonumber\]
    6. Find \[ \frac{d^3y}{dx^3}|_{x=1} \nonumber\] where \[ y=\frac{2}{x^3} \nonumber\].
    7. Suppose \[ u′(0)=98 \nonumber\] and \[(\frac{u}{q})′(0)=7 \nonumber\] Find q(0) assuming u(0)=0?
    8. Given: \[ b(x)=\frac{x^2−5x+4}{−5x+2} \nonumber\] What is: b′(2)?
    9. Given: \[ m(x)={e^x}{3x+4} \nonumber\] What is \[ \frac{dm}{dx} \nonumber\]?
    10. What is \[ \frac{d}{dx}⋅\frac{sin(x)}{x−4}? \nonumber\]
    11. Given \[ q(x)=\frac{x}{sin(x)} \nonumber\] What is \[ q′′(x)=\frac{d^2}{dx^2}\frac{x}{sin(x)}? \nonumber\]
    12. The position of a certain nano particle can be approximated by the function t3+t. What function gives the acceleration of the particle?
    13. The position of a car is given by the function sin(t)+3t2. Is the car accelerating or decelerating?
    14. The position of a velociraptor chasing a triceratops is given by the function cos(−t). Is the raptor experiencing positive or negative jerk at \[ t=\frac{3π}{2} \nonumber\]
    15. The position of the moon is night sky given by the following function of time: \[ \frac{1}{12}t^4−\frac{3}{6}t^3−5t^2+π^π \nonumber\] Name a time when the moon is experiencing no acceleration at all.
    16. What is the maximum number of times one would have to differentiate a N-degree polynomial before the derivative becomes zero?

    Review (Answers)

    To see the Review answers, open this PDF file and look for section 3.11.


    Term Definition
    Acceleration Acceleration is the rate at which the speed and direction of an object are changing.
    higher order derivative A higher order derivative is a second, or third, or nth derivative of a function.
    Instantaneous acceleration The instantaneous acceleration of an object is the change in velocity of the object calculated at a specific point in time.
    Instantaneous velocity The instantaneous velocity of an object is the velocity of the object at a specific point in time.

    Additional Resources

    PLIX: Play, Learn, Interact, eXplore - Higher Order Derivatives-Acceleration and Jerk

    Video: Calculus - Finding Higher Derivatives

    Real World: Deep Secrets

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