Thefourth derivative of a function is the derivative of its third derivative, the fifth derivative is the derivative of the fourth derivative, and so on. The nth derivative of a function is what we get when we differentiate it n times, where n is a positive integer. Consider y =f (x ). The nth derivative of f with respect to x can be written in various ways:
dny /dx n d nf(x ) /dxn d n/dx nf (x ) d nf /dx n
f (n) (x ) f (n)
Beyond the Third Derivative 133
As before, the parentheses around a superscript indicate that we’re talking about a derivative, not a power.
An example
Let’s work out the fourth derivative of the same three functions for which we have already found the second and third derivatives. The first original function, again, is
f (x ) = − 4x4+ 5x 3+ 6x 2− 7x+ 8 When we calculated the third derivative, we got
f ′′′(x ) = − 96x+ 30 Differentiating again, we obtain
f (4)(x ) = − 96
Another example
Now let’s find the fourth derivative of
f (x ) = 4x4− 5 cos x We found its third derivative to be
f ′′′(x ) = 96x− 5 sin x We must differentiate again! The sum rule tells us that
f (4)(x ) =d /dx (96x ) −d /dx (5 sin x ) Using the power rule and the multiplication-by-constant rule, we get
f (4)(x ) = 96 − 5 d /dx sin x The derivative of the sine is the cosine, so we have
f (4)(x ) = 96 − 5 cos x
Still another example
Now we’ll find the fourth derivative of
g (t ) = 2t 2+ 5t− 7 + 4t−1−t−2+ 2t−3+ ln t
We found that the third derivative is
g ′′′(t ) = 2t−3− 24t−4+ 24t−5− 120t−6 Using the generalized power rule and working term-by-term, we get
g (4)(t ) = −6t−4+ 96t−5− 120t−6+ 720t−7
Are you astute?
By now, you’ll have noticed that these three functions “morph” differently as you take their derivatives repeatedly. In the first case, if you differentiate again, you’ll get the zero function; after that, higher deriva-tives won’t change anything. In the second case, the constant −96 will vanish, leaving 5 or −5 times a sine or cosine. In the last case, you’ll always get a “nonstandard” polynomial. The exponents will get larger negatively, and the coefficients will alternate between positive and negative as their absolute values grow.
Can you tell, merely by looking at a function, what will occur if you differentiate it multiple times? As an extra-credit exercise, invent a few functions on your own. Then try to predict what will happen if you differentiate them over and over. Finally, do the calculations to see how accurate your predictions are.
Here’s a challenge!
Find the first nine derivatives of the sine function.
Solution
To work out all these derivatives, we need to know only that d /dx (sin x ) = cos x and
d /dx (cos x ) = − sinx Succeeding derivatives are easy to find:
d 2/dx 2 (sin x ) = d /dx (cos x ) = − sinx d 3/dx 3 (sin x ) = d /dx (− sinx ) = − cosx
d4/dx4 (sin x ) = d /dx (− cosx ) = sinx d 5/dx5 (sin x ) = d /dx (sin x ) = cosx d 6/dx6 (sin x ) = d /dx (cos x ) = − sinx d 7/dx7 (sin x ) = d /dx (− sinx ) = − cosx
d 8/dx8 (sin x ) = d /dx (− cosx ) = sinx d9/dx9 (sin x ) = d /dx (sin x ) = cosx
As we keep going, the derivatives cycle endlessly through these four functions in order: negative sine, nega-tive cosine, sine, and cosine.
Beyond the Third Derivative 135
Here’s another challenge!
Find the first nine derivatives of the natural log function, given that the domain is restricted to the posi-tive reals.
Solution
In this situation, we don’t get a cycle, but we will see a pattern. To begin, remember that d /dx (ln x ) =x−1
We can use the generalized power rule to find higher derivatives, building our results one upon another:
d 2/dx 2 (ln x ) = d/dx (x−1) = −x−2 d 3/dx 3 (ln x ) = d/dx (−x−2) = 2x−3 d 4/dx4 (ln x ) = d/dx (2x−3) = −6x−4 d 5/dx5 (ln x ) = d/dx (− 6x−4) = 24x−5 d 6/dx6 (ln x ) = d/dx (24x−5) = −120x−6 d 7/dx7 (ln x ) = d/dx (−120x−6) = 720x−7 d 8/dx8 (ln x ) = d/dx (720x−7) = −5,040x−8 d 9/dx9 (ln x ) = d/dx (−5,040x−8) = 40,320x−9
Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems.
Don’t hurry! You’ll find worked-out answers in App. A. The solutions in the appendix may not represent the only way a problem can be figured out. If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it!
1. In the chapter text, we found the second, third, and fourth derivatives of f (x ) = − 4x4+ 5x 3+ 6x 2− 7x+ 8
Find the fifth derivative of this function.
2. Find the sixth derivative of
f (x ) = − 4x4+ 5x 3+ 6x 2− 7x+ 8
3. In the chapter text, we found the second, third, and fourth derivatives of f (x ) = 4x4− 5 cos x
Find the fifth derivative of this function.
4. Find the sixth derivative of
f (x ) = 4x4− 5 cos x
5. Find the seventh derivative of
f (x ) = 4x4− 5 cos x
What happens as we keep on going to the eighth, ninth, and higher derivatives?
6. In the chapter text, we found the second, third, and fourth derivatives of g (t)= 2t 2+ 5t− 7 + 4t−1−t−2+ 2t−3+ ln t
Find the fifth derivative of this function.
7. Find the sixth derivative of
g (t ) = 2t 2+ 5t− 7 + 4t−1−t−2+ 2t−3+ ln t
8. Find the seventh derivative of
g (t ) = 2t 2+ 5t− 7 + 4t−1−t−2+ 2t−3+ ln t
What happens (in general) as we keep on going to the eighth, ninth, and higher derivatives?
9. Go back to the first “challenge” in the chapter text. Suppose Sir Isaac went to another cliff only 50 m high, and repeated the experiment. How long did the apple take to fall from that cliff ? What was the vertical component va of the apple’s speed at impact?
What was the vertical component aa of the apple’s acceleration at impact?
10. Look one more time at the first “challenge.” Imagine that, instead of working on earth, Sir Isaac conducted his experiment on a planet where the fallen distance h (in meters) as a function of time t (in seconds) was
h= 3t 2
If the apple took 11 seconds to fall, what was the height hc of the cliff ? What was the vertical component va of the apple’s speed at impact? What was the vertical component aa of the apple’s acceleration at impact?
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