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 denominators:

and the outermost variable, y, goes only in 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.
Practice:
Let z = (3x + 5)4. Find . | |
We can think of z as z = (y)4, where y = 3x + 5. Then 
Since z = y4, 
Since y = 3x + 5, .
Putting this information back into the chain rule, 
This is almost the answer, but since we're asked for we want to give the final answer in terms of x. We plug in 3x + 5 for y to find  That's all!
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Let z = cos(sin x). Find the derivative of z with respect to x. | |
Think of z as cos (y) where y = sin x. We want to find we use the chain rule. 
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If we have q = f(r) and r = g(i), what does the chain rule say? | |
The variable that keeps track of the { inside function} needs to be the one that "cancels out" in the chain rule: 
The input to the inside function, i, will occur only in denominators: 
and the outermost variable, q, will occur only in numerators: 
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Let u = 5z + z2 and let y = ln u. Find  | |
u is the inside function and z is the innermost variable, the form of the chain rule we want is 
The derivative of y = ln u with respect to u is 
and the derivative of u = 5z + z2 with respect to z is 5 + 2z. Applying the chain rule we find 
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Let q = sin(6s + s-1). Find . | |
This problem doesn't specify the inside function, so we'll make a name for it. Let y = 6s + s-1 and then q = sin(y). We want to use the version of the chain rule that says 
The derivative of q = sin(y) with respect to y is cos(y) and the derivative of y = 6s + s-1 with respect to s is 6 - s-2 We find this: 
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Answer
- We can think of z as z = ln y where y = x5 + 4x2.
Using the chain rule,

Answer
- We can think of z as z = y7 where y = sin x + cos x.
Using the chain rule,

Answer
- We can think of z as
where y = x9 - cos x.
Using the chain rule,

Answer
- We can think of z as z = ey where y = 12x + 4.Using the chain rule,

Answer
- There are two ways to do this.Way 1: We can think of z as z = ln y where
.Using the chain rule,

Way 2: First, rewrite the function:

Then taking the derivative only requires the rule for constant multiples, therefore

For the pair of functions, determine what the chain rule says.
- Let p = f(s) and s = g(z).
Answer
- p = f(s) and s = g(z), therefore p = f(s) = f({ g(z)}). s is the inside function, it will be the one that "cancels out":

The input to the inside function, z, will occur only in denominators:

and the outermost variable, p, will occur only in numerators:

The chain rule says

For the pair of functions, determine what the chain rule says.
- Let r = g(t) and q = f(r).
Answer
- The middle function is r, since q = f(r) = f(g(t)), therefore

The innermost variable is t:

and the outermost variable is q:

For the pair of functions, determine what the chain rule says.
- Let x = f(y) and y = g(z).
Answer
- Since x = f(y) and y = g(z), the overall function is
x = f(y) = f(g(z)).
The inner function is y, the innermost variable is z, and the outermost variable is x, the chain rule says

- If u = es and s = 5 - 3x2, find
.
Answer
- If u = es and s = 5 - 3x2,
the chain rule says

The derivative of u with respect to s is es and the derivative of s with respect to x is -6x, therefore we find

- If s = t4 and t = r4 + 1, find
.
Answer
- If s = t4 and t = r4 + 1, the chain rule says

The derivative of s with respect to t is 4t3 and the derivative of t with respect to r is 4r3, we find

- If m = sin n and z = ln m, find
.
Answer
- If m = sin n and z = ln m, the chain rule says

The derivative of z with respect to m is
and the derivative of m with respect to n is cos n, we find

- If z = (x4 + 4x) and y = z3 + z5, find
.
Answer
- If z = (x4 + 4x) and y = z3 + z5, the chain rule says

The derivative of y with respect to z is 3z2 + 5z4 and the derivative of z with respect to x is4x3 + 4, we find

- If
and q = tan p, find
.
Answer
- If
and q = tan p, the chain rule says

The derivative of r with respect to q is
and the derivative of q with respect to p is
sec2{p}, therefore we find

- Find
given that r = sin(x2 + 1) + cos(x2 + 1).
Answer
- Let y = x2 + 1 so that we can think of r as sin y + cos y.

- Find
given that p = e{r2 + 3r}
Answer
- Let y = r2 + 3r so that we can think of p as p = e{y}. Then

- Find
given that y = ln(z4 + z3)
Answer
- Let m = z4 + z3 so we can think of y as ln m. Then

- Find
given that
Hint
rewrite y to remove the fraction
Answer
- Let m = x2 + 4x - 2 so we can think of y as
. Then

- Find
given that q = 14{12sin t}.
Answer
- Let y = 12sin t and think of q as 14y. Then
