We stated the chain rule first in Lagrange notation. Since Leibniz notation lets us be a little more precise about what we're differentiating and what we're differentiating with respect to, we need to also be comfortable with the chain rule in Leibniz notation.
Suppose y is a function of x:
y = g(x)
and z is a function of y:
z = f(y)
Then z is a function of x:
z = f(y) = f(g(x))
Once again, we have an outside function and an inside function. The chain rule in Lagrange notation states that
(f(g(x))' = f ' (g(x)) · g ' (x).
In Leibniz notation, we would say
Since
and (f(g(x))' both mean "the derivative of z with respect to x,"
and f ' (g(x)) = f ' (y) both mean "the derivative of z with respect to y," and
and g ' (x) both mean "the derivative of y with respect to x,"
the two statements of the chain rule do mean the same thing.
We can remember the chain rule in Leibniz notation because it looks like a nice fraction equation where the dy terms cancel:
This may or may not be what's actually going on, but it works for our purposes and it's a great memory aid.
There are three steps to apply the chain rule in this form:
- determine what y is (this is the same step as determining the inside and outside functions)
- apply the chain rule formula
- put everything in terms of the correct variable (for example, writing y in terms of x)
The chain rule will be especially useful when we discuss related rates, where there will be problems with three different variables that all depend on each other in funny ways.
We can also use the chain rule with different letters, as long as we put the letters in the correct places. If we have
y = g(x) and z = f(y) = f(g(x)), the chain rule says
The inside function is the one that "cancels out":
The innermost variable, x, goes only in the denominators:
and the outermost variable, z, goes only in the numerators:
If we switch the letters, we need to make sure they go in the appropriate places.
The important thing is that the inside function needs to be the term that cancels out:
Now that we know how to write the chain rule with different letters, we can use it to find derivatives.
Example 1
Let z = y4 and y = 3x + 5. What is |
.
we want to give the final answer in terms of x. We plug in 3x + 5 for y to find
Example 2
Let z = cos(sin x). Find the derivative of z with respect to x. |
so we'll use the chain rule.
Example 3
If we have q = f(r) and r = g(i), what does the chain rule say? |
Example 4
Let u = 5z + z2 and let y = ln u. Find |
Example 5
Let q = sin(6s + s-1). Find |
.
Exercise 1
- If z = ln(x5 + 4x2), then what is the derivative of z with respect to x?
Exercise 2
- If z = (sin x + cos x)7 what is the derivative of z with respect to x?
Exercise 3
- What is the derivative of z with respect to x if
?
where y = x9 – cos x.
Exercise 4
- If z = e12x + 4 what is the derivative of z with respect to x?
Exercise 5
- If
, what is the derivative of z with respect to x?
.Using the chain rule,
Exercise 6
For the pair of functions, determine what the chain rule says.
- Let p = f(s) and s = g(z).
Exercise 7
For the pair of functions, determine what the chain rule says.
- Let r = g(t) and q = f(r).
Exercise 8
For the pair of functions, determine what the chain rule says.
- Let x = f(y) and y = g(z).
Exercise 9
- If u = es and s = 5 – 3x2, find
.
Exercise 10
- If s = t4 and t = r4 + 1, find
.
Exercise 11
- If m = sin n and z = ln m, find
.
and the derivative of m with respect to n is cos n. Putting it all together gives us
Exercise 12
- If z = (x4 + 4x) and y = z3 + z5, find
.
Exercise 13
- If
and q = tan p, find 
.
and q = tan p, the chain rule says
and the derivative of q with respect to p is 
Exercise 14
- What is
given that r = sin(x2 + 1) + cos(x2 + 1)?
Exercise 15
- What is
if p = er2 + 3r?
Exercise 16
- Find
given that y = ln(z4 + z3).
Exercise 17
- Find
given that 
.
. Then
Exercise 18
- Find
given that q = 1412sin t.
