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

h(x) = sin( x^{2})

We have used ` g(x) = Öx `` g(x) = sqrt(x)`

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 `x ^{2}`.
If we start out with:

_ g(x) = Öx eq. 4.4-1Then you can take the square of both sides to get

gYou do understand, don't you, that taking the square of the square root of^{2}(x) = x eq. 4.4-2

Let `f(x) = x ^{2}`

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``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. 4.4-4or by using the expression we have for

2g(x) g'(x) = 1 eq. 4.4-5Now all we need to do is get

1 g'(x) =Of course we'd still like to see an explicit expression for~~eq. 4.4-6 2g(x)~~

1 g'(x) =And there you have the derivative of~~eq. 4.4-7 2Öx~~

**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`

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

If you multiply numerator and denominator by

_____ _ (Öx + h + Öx )you get (by applying the

(x + h) - x g'(x) = limIn the numerator, the~~h~~~~> 0 h(Öx + h + Öx )~~

(In the end, you have a numerator of~~x~~+~~h~~) -~~x~~g'(x) = lim~~h~~~~> 0~~~~h~~(Öx + h + Öx )

_____ _ (Öx + h + Öx )As

1 g'(x) =just as we did before. Just goes to show that the chain rule really does work.~~eq. 4.4-7 2Öx~~

where `n` is an integer.
We take the same approach to this as to the previous problem. Remember that
`x ^{1/n}` is simply the

**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.

By now, you should be getting good at these chain rule problems. From
the previous coached exercise, you now know that the derivative of
`x ^{1/n}` is

xRemember 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.^{m/n}= (x^{1/n})^{m}eq. 4.4-13

**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.

**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 - x^{2}_________________ b) h(x) = Öx^{3}- 7x^{2}+ 3x - 4 c) h(x) = ( 1 - x^{2})^{3/2}

**4) Test your medal.** Rememeber that the derivative of `sin(x)` is
`cos(x)`. Recall also from trig that
` sin ^{2}(x) + cos^{2}(x) = 1 `

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) = sin ^{3}(x^{2})`

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:

- Square
`x` - Take the result of the previous step and take the
`sin`of it. - Take the result of the previous step and cube it.

**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(x ^{2})`

3 sinBut we aren't done yet. We still have two more steps of the "recipe" to cover.^{2}(x^{2})

**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 `x ^{2}`. So accordingly
we take

3 sinBut we are still not done. We still have the first step of the "recipe" to cover.^{2}(x^{2}) cos(x^{2})

**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 sinwhich is indeed the derivative of^{2}(x^{2}) cos(x^{2}) (2x)

u'(x) = 6x sinbut doing that is just window dressing.^{2}(x^{2}) cos(x^{2})

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.

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

Even though we don't know exactly what^{ }____ y^{2}(x) + x^{3}= Öy(x) eq. 4.4-20

We can see that the first term, `y ^{2}(x)` is the composite
of

The next term in the equation is `x ^{3}`. We know that its
derivative is

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

1Gathering all this into one equation, we have~~y'(x) 2Öy(x)~~

1 (2y(x) y'(x) ) + 3xThis doesn't give you an exact expression for^{2}=~~y'(x) eq. 4.4-21 2Öy(x)~~

Let's try another implicit differentiation problem.

sin( y(x) ) sin(x) = 1 eq. 4.4-22You will recall that in a previous section I tipped you off that the derivative of the

(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

and its solution would appear^{ }_ y^{2}+ x^{3}= Öy eq. 4.4-24

1 (2y y') + 3x^{2}=~~y' eq. 4.4-25 2Öy~~

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`).

_ yAgain 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^{2}Öx = y eq. 4.4-26

1respectively. Now applying the product rule to the left side we have~~2Öx~~

y^{2}(2y y' sqrt(x) ) +~~= y' eq. 4.4-27 2Öx~~

Let's try

______ Ö1 - yThis one has the added wrinkle that you have nested composites. That is, you have square root on the outside and^{2}= x^{2}eq. 4.4-28a

Let ` f(g) = Ö g`` g(x) = 1 - x ^{2}`

______ Ö1 - yWe know that^{2}= f( g(y) ) = x^{2}eq. 4.4-28b

1 f'(x) =and that~~2Öx~~

f'( g(y) ) g'(y) = f'( g(y) ) (-2y y') eq. 4.4-29aor, 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 - y~~^{2}

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

xNote the wording. Think about what is a function of what. We have to apply the chain rule to both^{2}+ y^{2}= R^{2}t eq. 4.4-30

Taking the derivative of the right hand side of the equal is easy. It
was given that `R` is a constant, so `R ^{2}` is
also a constant. We know that

When we put that all together, we get

(2x x') + (2y y') = R^{2}eq. 4.4-31

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)where^{3}______ b) h(x) = Ö3x + 4 _______ c) h(x) = Öx^{3}+ 4x d) h(x) = ( x^{2}+ 2x + 1 )^{3/2}e) h(x) = ( 1 - (2x + 1)^{2})^{3/2}^{ }______ f) h(x) = ( a^{2}- Öx^{2}+ 1 )^{2/3}

**6)** Given that `e ^{x}` and

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

where^{ }_ a) y^{3}+ x^{2}= Öx

b) xy = 1where

_______ c) Öxwhere^{2}+ y^{2}= R

x + y d)where~~= 1 2x - y~~

e) uwhere^{1/2}v^{1/3}= (u + v) t^{2}

f) (uwhere^{-1/2}+ v^{-1/3})^{2}= t^{2}

**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)h ^{2}` liters. In other words,

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.

- 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.
- At what time (in hours relative to now) will their rate of change of distance be zero?
- 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
`x`is the east-west position of Ship A and_{A}`y`is the north-south position of Ship A, and likewise_{A}`x`and_{B}`y`are the east-west and north-south coordinates of Ship B, then we can say that_{B}`x`(that is the derivative of_{A}'`x`with respect to time) is the east-west speed and_{A}`y`is the north-south speed (in knots) of Ship A. Likewise with_{A}'`x`and_{B}'`y`for Ship B. Notice that all these variables are functions of the independent variable of time,_{B}'`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(x

_{B}- x_{A})(x_{B}' - x_{A}') + (y_{B}- y_{A})(y_{B}' - y_{A}') = 0

The Soft Touch (finding tangent lines)

email me at *hahn@netsrq.com*