Section 4: DerivativesKCT logo

© 1996 by Karl Hahn

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4.4 Chain Rule Applications

The chain rule is admittedly the most difficult of the rules we have encountered so far. But it is also the most powerful. You must be able to apply the mechanics of this rule before you will be ready for the next challenge, which is knowing when to apply it. This is because the chain rule's usefulness goes beyond the problem of finding the derivative of something that is explicitly the composite of two functions. Throughout calculus we will be making substitutions of variables. It happens all the time. And every time we do, the chain rule will apply. And there are other applications, as we shall see shortly.

So before proceding with this section, be sure that you understand the statement of the chain rule and the example that follows it. Review it until you have some confidence in your grasp of it. If reviewing the story about the professor's watch helps, then review that as well. Then come back here and see if you can do the exercises that follow. When you can do the exercises, then procede to what follows them.


1) Apply the chain rule to find the derivatives of the following. Check your work by expanding the expression shown below and using other methods we have learned to arrive at the same answer.

a) h(x) = (x + 1)2

b) h(x) = (2x - 1)2 - 3(2x - 1) + 2
2) Although we haven't covered trig functions yet, I will tell you that if  g(x) = sin(x)  then  g'(x) = cos(x)  (proof of this will be furnished in a later section). Given this relationship, find the derivative of
   h(x) = sin( x2 )

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Finding the Derivative of an Inverse Function

We have used  g(x) = Ö (that's the same as  g(x) = sqrt(x)) in several examples so far. And I even mentioned that some instructors might have you use a bastardized version of the binomial theorem to find its derivative. But there is an easier and more persuasive way to find this derivative, and it involves the chain rule.

Recall that if you do anything except divide by zero to both sides of an equation, you still have a valid equation, as long as what you did was the same on both sides of the equals. This includes taking a function. We know that Öx is the inverse function of x2. If we start out with:

   g(x)  =  Öx                                                   eq. 4.4-1
Then you can take the square of both sides to get
   g2(x)  =  x                                                   eq. 4.4-2
You do understand, don't you, that taking the square of the square root of x always returns the original x because square and square root are, by definition, inverse functions of each other.

Let f(x) = x2. Then we have

   f( g(x) )  =  x                                               eq. 4.4-3

We are now in a position to take the derivative of both sides of 4.4-2. We know that the derivative of x is 1. That takes care of taking the derivative of the right hand side. And the left hand side is a composite, so we can apply the chain rule. We know that  f'(x) = 2x. We don't know g'(x) yet -- in fact that is what we are trying to find out. So by applying the chain rule to 4.4-3, we have

   f'( g(x) ) g'(x)  =  1                                        eq. 4.4-4
or by using the expression we have for f'(x), we can substitute and get
   2g(x) g'(x)  =  1                                             eq. 4.4-5
Now all we need to do is get g'(x) all by itself on one side of the equal. And we can do that by dividing out 2g(x).
   g'(x)  =                                                      eq. 4.4-6
Of course we'd still like to see an explicit expression for g'(x), but we can do that because we already know what g(x) is from equation 4.4-1. We simply substitute g(x) from that equation.
   g'(x)  =                                                      eq. 4.4-7
And there you have the derivative of Öx, done by applying some algebra and the chain rule.

Crosschecking by taking the limit: Just to check that we can come up with the same answer using the limit method, observe that the derivative of g(x) = Öx is given by

                       _____    _
                      Öx + h - Öx
   g'(x)  =   lim                
             h  > 0        h
according to our original definition of what a derivative is.

If you multiply numerator and denominator by

     _____    _
   (Öx + h + Öx )
you get (by applying the difference of squares):
                      (x + h) - x
   g'(x)  =   lim                  
             h  > 0 h(Öx + h + Öx )
In the numerator, the x's cancel. Then the h on the left of the denominator cancels with the h remaining in the numerator.
                        (x + h) - x
   g'(x)  =   lim                    
             h  > 0   h(Öx + h + Öx )
In the end, you have a numerator of 1 and a denominator of
     _____    _
   (Öx + h + Öx )
As h approaches zero, Ö(x + h) approaches Öx. So when you take the limit, you end up with
   g'(x)  =                                                      eq. 4.4-7
just as we did before. Just goes to show that the chain rule really does work.

Coached Exercise: Find the derivative of x1/n

where n is an integer. We take the same approach to this as to the previous problem. Remember that x1/n is simply the nth root of x, and the nth root of x is simply the inverse function of xn.

Step 1: Write let g(x) be the function we are interested in finding the derivative and let f(g) be its inverse. Write equations for both of these. Label them 4.4-8a and 4.4-8b respectively. You should be able to write the expression for f'(g) as well. Label that equation 4.4-8c.

Step 2: Take the composite of the two functions. That is equal to what? Remember that a composite of two functions that are inverses of each other is always equal to the same thing. Do you remember what that is? Write the composite (using your f and g symbols) on the left of the equal, with the function we are interested in on the inside of the composite, and its inverse on the outside. Write the expression that represents what the composite of a function with its inverse always is on the right of the equal sign. Label this equations 4.4-9.

Step 3: Take the derivative of both sides of equation 4.4-9. On the left, you have a composite, so you apply the chain rule. The expression on the right should be easy to take the derivative of. Label your result 4.4-10.

Step 4: Substitute back. In equations 4.4-8a, 4.4-8b, and 4.4-8c, you have expressions for f(x), f'(x), and g(x). Substitute back for f'(x) first. Label that 4.4-11. Then substitute back for g(x). Label that 4.4-12.

Step 5: Solve for g'(x). You do this with just a little algebra.

If you are confused, go back and review how we did the same problem when we found the derivative of sqrt(x). The reason I say it is the same problem is because it is, only in that one we have set  n = 2.

You can go to the solution by clicking here, but please, not until you have made a sincere effort to solve this problem on your own.

Another Coached Exercise: Derivative of xm/n

By now, you should be getting good at these chain rule problems. From the previous coached exercise, you now know that the derivative of x1/n is  (1/n) x(1-n)/n, where n is an integer. Now, keeping that result in mind, can you use the chain rule to find the derivative of xm/n, where m and n are both integers? Recall from algebra that

   xm/n  =  (x1/n)m                                             eq. 4.4-13
Remember that taking a power of a power is the same as multiplying the exponents. Notice that right hand side of 4.4-13 is the composite of two functions, which means that the chain rule ought to get us its derivative.

Step 1: What are the two functions that the right hand side of 4.4-13 is the composite of? Call the inner one g(x) and the outer one f(g). Write equations for both, and label them 4.4-14a and 4.4-14b respectively.

Step 2: Find the derivatives of f(g) and g(x). Label them 4.4-15a and 4.4-15b respectively.

Step 3: Let's call the composite function h(x). Write an equation for h(x) as a composite using your f and g symbols. Label this equation 4.4-16.

Step 4: Apply the chain rule to to find h'(x) in terms of your f and g symbols. Label this equation 4.4-17.

Step 5: Substitute back into 4.4-17 from 4.4-14a, 4.4-15a and 4.4-15b. You may want to do this in several stages.

Step 6: Use some algebra to simplify the expression that ended up with in step 5.

If you are still confused about the use of the chain rule, go back and review it from the start. You must get comfortable with applying this rule because it will come up again and again in your later studies.

Again, you can see the solution by clicking here. But again, do please make a sincere effort before you do so.

More Chain Rule Exercises

3) Use the chain rule and the formulae you learned in this section for derivatives of fractional powers to find the derivatives of the following:

a)  h(x) = Ö1 - x2
b)  h(x) = Öx3 - 7x2 + 3x - 4

c)  h(x) = ( 1 - x2 )3/2

4) Test your medal. Rememeber that the derivative of sin(x) is cos(x). Recall also from trig that  sin2(x) + cos2(x) = 1  for all x. Using a method similar to what we used to find the derivative of Öx, find the derivative of the inverse function of sin(x). This function is commonly denoted either arcsin(x) or sin-1(x). Your approach will start with letting  g(x) = sin-1(x)  and let  f(g) = sin(g) . Then let h(x) be the composite of these two, with f on the outside and g on the inside. I'll let you take it from there.

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Applying the Chain Rule to Longer Chains

Some students, even when they understand how to apply the chain rule to composites of two functions (that is f(g(x))), still have difficulty when the instructor confronts them with composites of three or more functions. As an example, we shall apply the chain rule here to find the derivative of  u(x) = sin3(x2).

I'd like you to think of the u(x) given above as a recipe. This recipe tells you to take whatever x is given and apply certain operations to it in a particular order. The order is established by taking it from the inside out. Here is the recipe that u(x) calls for:

  1. Square x
  2. Take the result of the previous step and take the sin of it.
  3. Take the result of the previous step and cube it.
To take the derivative of u(x), we will have to use the chain rule twice. Why? Because on a composite of two functions you apply the chain rule once. This is a composite of three, so you apply it twice (always the number of compositions is one more than the number of applications of the chain rule required). What's more, we will apply the chain rule while running through the "recipe" backward. Here goes:

Step 1: The last step of the "recipe" says to take the cube of something. The derivative of taking the cube is taking the square and multiplying by 3. So we do that to everything the recipe takes the cube of. And what we are taking the cube of is sin(x2). So we take 3 times the square of that:

   3 sin2(x2)
But we aren't done yet. We still have two more steps of the "recipe" to cover.

Step 2: The next to last step of the "recipe" says to take the sin of something. The derivative of taking the sin is taking the cos. So we take everything we were taking the sin of and take the cos of it instead. And then we multiply that by what we got in step 1. Observe that u(x) calls for us to take the sin of x2. So accordingly we take cos(x2) and multiply that by what we got in step 1:

   3 sin2(x2) cos(x2)
But we are still not done. We still have the first step of the "recipe" to cover.

Step 3: The first step of the "recipe" says to square x. The derivative of squaring x is multiplying x by 2. That gives 2x. And we multiply that by what we got in step 2:

   3 sin2(x2) cos(x2) (2x)
which is indeed the derivative of u(x). Of course you can reorganize the terms to make it look nicer:
   u'(x)  =  6x sin2(x2) cos(x2)
but doing that is just window dressing.

If you ever get confused on a problem like this one where there is a composite of three or more functions, try doing it just as I've done here. Write out the recipe, then go through it backward. On each step of the recipe, ask yourself, "What is the derivative of this step?" Then apply that derivative to everything the recipe's step is applied to. Then do the same with the next prior step and multiply what you get with what you've already got. Keep repeating that process until you've covered the entire recipe from bottom to top.

Implicit Differentiation (putting it all together)

Supposing we have a function, y(x), and we don't know exactly what it is. We know something about it, because we have an equation that it fits into, but solving that equations for y(x) would be difficult. For example suppose we have

   y2(x) + x3  =  Öy(x)                                      eq. 4.4-20
Even though we don't know exactly what y(x) is, we can still use the chain rule to find out something about its derivative. Lets take the derivative of both sides of 4.4-20, but do it step by step.

We can see that the first term, y2(x) is the composite of y(x) on the inside and x2 on the outside. Applying the chain rule to that, we see that the derivative of that term is 2y(x) × y'(x) (note that we have come far enough with the chain rule now that you should be able to apply it without having to explicitly break it into an f(x) and a g(x). You ought to be able to apply the chain rule by inspection now).

The next term in the equation is x3. We know that its derivative is 3x2.

On the right hand side of 4.4-20 we have another composite, Öy(x) (that is sqrt(y(x))). Again we can apply the chain rule to find that this term's derivative is

Gathering all this into one equation, we have
   (2y(x) y'(x) ) + 3x2  =         y'(x)                          eq. 4.4-21
This doesn't give you an exact expression for y'(x), but 4.4-20 didn't give you an exact expression for y(x) either. But it does tell you something about y'(x), and besides, this type of problem is likely to appear on your exam.

Let's try another implicit differentiation problem.

   sin( y(x) ) sin(x)  =  1                                      eq. 4.4-22
You will recall that in a previous section I tipped you off that the derivative of the sin is the cos. Here we have a product on the left, but one of the factors in that product is a composite of sin and y(x). So we will have to apply both the chain rule and the product rule. You didn't forget the product rule, did you? Applying the chain rule to find the derivative of the left factor of the product, we find its derivative to be  cos( y(x) ) y'(x). The derivative of the other factor we know to be cos(x). And the right hand side of the equation is constant, and you will recall that the derivative of a constant is always zero. So putting those expressions together with the product rule, we have
   (cos( y(x) ) y'(x) sin(x) )  +  (sin( y(x) ) cos(x) )  =  0

                                                                 eq. 4.4-23

In many if not most texts, they will leave the "(x)" out and just say "y is a function of x." This is just a change in notation, and it is nothing to be troubled over. In this form, the problem given in 4.4-20 would appear

   y2 + x3  =  Öy                                                eq. 4.4-24
and its solution would appear
   (2y y') + 3x2  =      y'                                      eq. 4.4-25

I'd like you to get used to working problems in this notation, since you will likely have to do them in your classwork this way. When you encounter such problems, look in the text, which will usually tell you what is a function of what. In this case we had y as a function of x, which means that you can imagine any occurrence of y in the problem as being y(x). If the text says that x is a function of t, then each time you saw x, you would imagine it as x(t). Often u and v are used as symbols for functions of either x or t. In those cases you would imagine u(x) or v(x) or u(t) or v(t) whenever you saw u or v. In polar coordinate problems (which we will get to in a later section), often r will be a function of theta.

In the case of y being a function of x, that usage is so common and conventional that problems are often given in that form without ever stating that y is a function of x. In that case, you may assume that it is. Of course, the same rule applies to y', which you may assume means y'(x). And remember that the independent variable (that is, the variable that appears inside the parentheses, in this case, x) will never have a ' after it. Only a function can have a derivative.

Here are a few more worked implicit differentiation examples (in which y is a function of x).

   y2 Öx  =  y                                                   eq. 4.4-26
Again we use both the chain rule and the product rule on the left side of the equal, and on the right, we know that the derivative of y is simply y'. The derivatives of the two factors on the left are 2y × y' (which we get by applying the chain rule) and
respectively. Now applying the product rule to the left side we have
   (2y y' sqrt(x) ) +      =  y'                                 eq. 4.4-27

Let's try

   Ö1 - y2  =  x2                                                eq. 4.4-28a
This one has the added wrinkle that you have nested composites. That is, you have square root on the outside and  1 - y 2 on the inside, but inside of that you have that y is a function of x. This means you'll have to apply the chain rule twice. We just have to take it one step at a time.

Let  f(g) = Ö g and let  g(x) = 1 - x2. Then this problem becomes

   Ö1 - y2  =  f( g(y) )  =  x2                                  eq. 4.4-28b
We know that
   f'(x)  =     
and that  g'(x) = -2x. But what is g'(y)? Well, we have another composite here, and so we apply the chain rule again:  g'(y) = -2y y'. So putting together both applications of the chain rule we have
   f'( g(y) ) g'(y)  =  f'( g(y) ) (-2y y')                      eq. 4.4-29a
or, substituting and including the derivative of the right hand side of 4.4-28a we have
                            -2y y'
   f'( g(y) ) (-2y y')  =            =  2x                       eq. 4.4-29b
                           2Ö1 - y2

Here's a curve ball that an instructor might throw you on an exam. Suppose that both x and y are functions of t, and R is a constant. Do the implicit differentiation on

   x2 + y2  =  R2t                                               eq. 4.4-30
Note the wording. Think about what is a function of what. We have to apply the chain rule to both x2 and to y2 because they are both functions of the independent variable, t. So the derivative of x2 is 2x x' and the derivative of y2 is 2y y'.

Taking the derivative of the right hand side of the equal is easy. It was given that R is a constant, so R2 is also a constant. We know that t is the independent variable, and that the derivative of t is always 1. Do you remember the rule about taking the derivative of constant times any expression? The rule is "the derivative of a constant times bananas is the same constant times the derivative of bananas" (where bananas is any expression that has a derivative). So the derivative of R2t is simply R2.

When we put that all together, we get

   (2x x') + (2y y')  =  R2                                     eq. 4.4-31

Yet More Exercises

You may have noticed that the problems we have been tackling lately have gotten a lot harder than stuff we did in earlier sections. Sorry, I can't help that. You may want to review part or all the preceding section several times before diving into these. This series will progress from fairly easy applications of the chain rule to more and more difficult ones. In the end, you should be able to do them all. When you can, you will know that you have mastered this material.

5) Apply the chain rule to find the derivatives of the following functions:

a)   h(x)  =  (3x + 4)3
b)   h(x)  =  Ö3x + 4
c)   h(x)  =  Öx3 + 4x

d)   h(x)  =  ( x2 + 2x + 1 )3/2

e)   h(x)  =  ( 1 - (2x + 1)2 )3/2
f)   h(x)  =  ( a2 - Öx2 + 1 )2/3
where a is a constant.

6) Given that ex and ln(x) are inverse functions of each other, and given that ex is its own derivative, use the method we used for finding the derivative of sqrt(x) to find the derivative of ln(x) (by the way, in a later section we will prove all the things I just said about ex and ln(x)).

7) Apply implicit differentiation to the following according to ground rule given for each example:

a)   y3 + x2  =  Öx
where y is a function of x.
b)   xy  =  1
where y is a function of x.
c)   Öx2 + y2  =  R
where y is a function of x and R is a constant.
      x + y
d)           =  1
     2x - y
where y is a function of x. Hint: you can use the
quotient rule to do this or you can use some algebra to make the implicit differentiation easier. Try it both ways and see that you come up with answers that are equivalent.
e)   u1/2v1/3  =  (u + v) t2
where u and v are both functions of t.
f)   (u-1/2 + v-1/3)2  =  t2
where u and v are both functions of t.

8 A certain vase has a strange shape. The volume, v, of water that it holds when filled to a height of h centimeters is (1/2)h2 liters. In other words, v(h) = (1/2)h2.

Given that the vase has a round cross section at every height, and given that the derivative (that is the rate of change) of volume with respect to height of water is always equal to the cross sectional area at that height, find the radius, r of the vase as a function of height, h.

9) Here is one that I have been asked about so many times by email that it's time to put it online. Two ships are steaming along on the ocean. They have the colorful names of Ship A and Ship B. Since this is a nautical problem, I'll use the nautical units for distance and speed, which are nautical miles for distance and knots for speed. One nautical mile is about 1850 meters. One knot is one nautical mile per hour. Right now Ship A is 20 nautical miles south of Ship B. Ship A is cruising east at 15 knots. Ship B is cruising south at 25 knots. All velocities remain constant.

  1. At what rate of speed (in knots) is their distance changing? The answer should be negative if they are closing on each other, positive if they are getting farther away.
  2. At what time (in hours relative to now) will their rate of change of distance be zero?
  3. Now here's a special challenge. Let's look at the more general case, where the two ships' positions and velocities are not given outright but as variables. If xA is the east-west position of Ship A and yA is the north-south position of Ship A, and likewise xB and yB are the east-west and north-south coordinates of Ship B, then we can say that xA' (that is the derivative of xA with respect to time) is the east-west speed and yA' is the north-south speed (in knots) of Ship A. Likewise with xB' and yB' for Ship B. Notice that all these variables are functions of the independent variable of time, t, which is given in hours. Can you show that the rate of change of distance between Ship A and Ship B is always zero whenever
       (xB - xA)(xB' - xA') + (yB - yA)(yB' - yA')  =  0
Hint: Set this up on a Cartesian grid with the x direction being east-west (east is positive) and the y direction being north-south (north is positive). Place Ship A at the origin. Use the symbols given in part c for the positions and velocities of the two ships. Use the Pythagorean formula for distance and take its derivative with respect to time using the
chain rule. The standard symbol to use for distance is s, so in this problem, s is a function of time, t, and it has its own derivative with respect to time, s'. The problem is asking you various things about s'. Part a is asking you to evaluate it at t = 0. Part b is asking you to find the t that makes s' = 0. Part c is asking you to prove that a certain configuration of relative position and relative velocity always leads to s' = 0. This last part is admittedly the hardest, but it teaches you to solve problems symbolically rather than numerically. As you get farther into calculus that becomes a more and more important skill. So give all three parts, including the last, your best effort.

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