7
THE CHAIN RULE
The derivative of a function of a function
The derivative of a function of a function
Let
f (x) = x5 and g(x) = x² + 1.
Then
f (g(x)) = (x² + 1)5.
What is the derivative of f (g(x)) ?
First, note that
| d f(x) dx |
= 5x4. |
That is: The derivative of f with respect to its argument (which in this case is x) is equal to 5 times the 4th power of the argument.
This means that if g -- or any variable -- is the argument of f, the same form applies:
| d f(g) dg |
= 5g4. |
| d f(h) dh |
= 5h4. |
| d f(v) dv |
= 5v4. |
And so on. Specifically, then, since g = x² + 1,
| d f(g) dg |
= 5g4 | = 5(x² + 1)4. |
What the chain rule states is the following:
| df(g(x)) dx |
= | df(g) dg |
· | dg dx |
"If f is a function of g and g is a function of x,
then the derivative of f with respect to x
is equal to the derivative of f with respect to g
times the derivative of g with respect to x."
Now, the derivative of g is 2x. Therefore here is the derivative of
(x² + 1)5.
5(x² + 1)4· 2x.
Note: In (x² + 1)5, x² + 1 is "inside" the 5th power, which is "outside." We take the derivative from outside to inside. When we take the outside derivative, we do not change what is inside. Then, we multiply by the derivative of what is inside.
When we write f(g(x)), g(x) is inside f. We take the derivative of f first.
| Example 1. f(x) = |  What is its derivative? |
Solution. x4 − 2 is inside the square root function. The derivative of the square root is given in the Example of Lesson 6. For any argument g of the square root function,
Here, g is x4 − 2. Therefore, since the derivative of x4 − 2 is 4x3,
| d dx |
|
= ½(x4 − 2)−½· 4x3 = 2x3(x4 − 2)−½. |
Example 2. What is the derivative of y = sin3x ?
Solution. sin3x is the 3rd power of sin x. sin x is "inside" the 3rd power. The derivative of the 3rd power -- of g3 -- is 3g². Therefore, accepting for the moment that the derivative of sin x is cos x (Lesson 12), the derivative of sin3x is
3 sin²x· cos x.
| Example 3. What is the derivative of | 1 x3 + 1 |
? |
| Solution. x3 + 1 is inside the function | 1 x |
= x−1, whose derivative |
is −x−2 ; (Problem 3, Lesson 4). Thus,
| 1 x3 + 1 |
= | (x3 + 1)−1 | . |
Therefore, its derivative is
−(x3 + 1)−2· 3x²
Example 4. Assuming that y is a function of x, apply the chain rule to
| express | d dx |
y² | . |
| Solution. 2y | dy dx |
. |
y is "inside" the function y².
Problem 1. Calculate the derivative of (x² −3x + 5)9.
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Do the problem yourself first!
9(x² −3x + 5)8(2x − 3)
Problem 2. Calculate the derivative of (x4 − 3x² + 4)2/3.
2/3(x4 − 3x² + 4)−1/3(4x3 − 6x)
Problem 3. Calculate the derivative of sin5x.
5 sin4x cos x
Problem 4. Calculate the derivative of sin x5.
The inside function is x5. The outside function is sin x. (This is the sine of x5.) Therefore, the derivative is
cos x5· 5x4.
Problem 5. Calculate the derivative of sin (1 + 2
).
cos (1 + 2
)x−1/2
Problem 6. Calculate the derivative of 
¼(sin x)−3/4 cos x
| * | * | * |
The chain rule can be extended to more than two functions. For example, let
| f(x) | = |  |
. |
The outside function is the square root. Inside that is (1 + a 2nd power). And inside that is sin x.
The derivative therefore is
| ½(1 + sin²x)−1/2· 2 sin x· cos x | = | sin x cos x |
| Problem 7. Calculate the derivative of | _ 1 _ sin (x² + 5) |
. |
(Compare Example 3.)
| −[sin (x² + 5)]−2· cos (x² + 5)· 2x | = | − |
2x cos (x² + 5) sin²(x² + 5) |
| Problem 8. Calculate the derivative of |  |
 |
Problem 9. Assume that y is a function of x, and apply the chain rule to express each derivative with respect to x.
| a) | d dx |
y3 = | 3y² | dy dx |
| b) | d dx |
sin y = | cos y | dy dx |
| c) | d dx |
 = |
½y−½ | dy dx |
Proof of the chain rule
To prove the chain rule let us go back to basics. Let f be a function of
g(x): f(g(x)). Then when x changes by an amount Δx, f will also change by an amount Δf. Hence the derivative of f with respect to x is the limit as Δx approaches 0, of
| Δf Δx |
. |
(Lesson 4. Whether we call the change Δf or Δy does not matter.)
Similarly, the change in f with respect to g is
| Δf Δg |
. |
| When Δg approaches 0, that becomes | df dg |
. |
The change in g with respect to x is
| Δg Δx |
. |
Here is the product of those quotients:
| Δf Δg |
· | Δg Δx |
. |
| This is equal to | Δf Δx |
: |
| Δf Δx |
= | Δf Δg |
· | Δg Δx |
. |
Now let Δx approach 0. Then Δg will approach 0 (Note, Lesson 4). Therefore, since the limit of a product is equal to the product of the limits (Lesson 2):
| df dx |
= | df dg |
· | dg dx |
This is the chain rule.
Next Lesson: The quotient rule
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