# Chain rule for differentiation

ORIGINAL FULL PAGE: Chain rule for differentiation
STUDY THE TOPIC AT MULTIPLE LEVELS:
ALSO CHECK OUT: Quiz (multiple choice questions to test your understanding) |Page with videos on the topic, both embedded and linked to
View other differentiation rules

## Statement for two functions

The chain rule is stated in many versions:

Version type Statement
specific point, named functions Suppose $f$ and $g$ are functions such that $g$ is differentiable at a point $x = x_0$, and $f$ is differentiable at $g(x_0)$. Then the composite $f \circ g$ is differentiable at $x_0$, and we have:
$\! \frac{d}{dx}[f(g(x))]|_{x = x_0} = f'(g(x_0))g'(x_0)$
generic point, named functions, point notation Suppose $f$ and $g$ are functions of one variable. Then, we have
$\! \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)$ wherever the right side expression makes sense.
generic point, named functions, point-free notation Suppose $f$ and $g$ are functions of one variable. Then,
$\! (f \circ g)' = (f' \circ g) \cdot g'$ where the right side expression makes sense, where $\cdot$ denotes the pointwise product of functions.
pure Leibniz notation Suppose $u = g(x)$ is a function of $x$ and $v = f(u)$ is a function of $u$. Then,
$\frac{dv}{dx} = \frac{dv}{du}\frac{du}{dx}$
MORE ON THE WAY THIS DEFINITION OR FACT IS PRESENTED: We first present the version that deals with a specific point (typically with a $\{ \}_0$ subscript) in the domain of the relevant functions, and then discuss the version that deals with a point that is free to move in the domain, by dropping the subscript. Why do we do this?
The purpose of the specific point version is to emphasize that the point is fixed for the duration of the definition, i.e., it does not move around while we are defining the construct or applying the fact. However, the definition or fact applies not just for a single point but for all points satisfying certain criteria, and thus we can get further interesting perspectives on it by varying the point we are considering. This is the purpose of the second, generic point version.

### One-sided version

A one-sided version of sorts holds, but we need to be careful, since we want the direction of differentiability of $f$ to be the same as the direction of approach of $g(x)$ to $g(x_0)$. The following are true:

Condition on $g$ at $x_0$ Condition on $f$ at $g(x_0)$ Conclusion
left differentiable at $x_0$ differentiable at $g(x_0)$ The left hand derivative of $f \circ g$ at $x_0$ is $f'(g(x_0))$ times the left hand derivative of $g$ at $x_0$.
right differentiable at $x_0$ differentiable at $g(x_0)$ The right hand derivative of $f \circ g$ at $x_0$ is $f'(g(x_0))$ times the right hand derivative of $g$ at $x_0$.
left differentiable at $x_0$, and increasing for $x$ on the immediate left of $x_0$ left differentiable at $g(x_0)$ the left hand derivative is the left hand derivative of $f$ at $g(x_0)$ times the left hand derivative of $g$ at $x_0$.
right differentiable at $x_0$, and increasing for $x$ on the immediate right of $x_0$ right differentiable at $g(x_0)$ the right hand derivative is the right hand derivative of $f$ at $g(x_0)$ times the left hand derivative of $g$ at $x_0$.
left differentiable at $x_0$, and decreasing for $x$ on the immediate left of $x_0$ right differentiable at $g(x_0)$ the left hand derivative is the right hand derivative of $f$ at $g(x_0)$ times the left hand derivative of $g$ at $x_0$.
right differentiable at $x_0$, and decreasing for $x$ on the immediate right of $x_0$ left differentiable at $g(x_0)$ the right hand derivative is the left hand derivative of $f$ at $g(x_0)$ times the left hand derivative of $g$ at $x_0$.

## Statement for multiple functions

Suppose $f_1,f_2,\dots,f_n$ are functions. Then, the following is true wherever the right side makes sense:

$(f_1 \circ f_2 \circ f_3 \dots \circ f_n)' = (f_1' \circ f_2 \circ \dots \circ f_n) \cdot (f_2' \circ \dots \circ f_n) \cdot \dots \cdot (f_{n-1}' \circ f_n) \cdot f_n'$

For instance, in the case $n = 3$, we get:

$(f_1 \circ f_2 \circ f_3)' = (f_1' \circ f_2 \circ f_3) \cdot (f_2' \circ f_3) \cdot f_3'$

In point notation, this is:

$\! \frac{d}{dx}[f_1(f_2(f_3(x)))] = f_1'(f_2(f_3(x))f_2'(f_3(x))f_3'(x)$

## Reversal for integration

If a function is differentiated using the chain rule, then retrieving the original function from the derivative typically requires a method of integration called integration by substitution. Specifically, that method of integration targets expressions of the form:

$\int h(g(x))g'(x) \, dx$

The $u$-substitution idea is to set $u = g(x)$ and obtain:

$\int h(u) \, du$

We now need to find a function $f$ such that $f' = h$. The integral is $f(u) + C$. Plugging back $u = g(x)$, we obtain that the indefinite integral is $f(g(x)) + C$.

## Significance

### Why more naive chain rules don't make sense

There are two naive versions of the chain rule one might come up with, neither of which holds:

$(f \circ g)'(x) = f'(g'(x))$

and

$(f \circ g)'(x) = f'(x)g'(x)$

Even without doing any mathematics, we can deduce that neither of these rules can be correct. How? Any rule that holds generically must involve evaluating $f$ or $f'$ only at points that we know to be in the domain of $f$. The only such point in this context is $g(x)$. Therefore, the chain rule cannot involve evaluating $f$ or $f'$ at any point other than $g(x)$. Note that our actual chain rule: [itex](f \circ g)'(x) = f'(g(x))g'(x)$ is quite similar to the naive but false rule $\! (f \circ g)'(x) = f'(x)g'(x)$, and can be viewed as the corrected version of the rule once we account for the fact that $f'$ can only be calculated after transforming $x$ to $g(x)$. ### Qualitative and existential significance Each of the versions has its own qualitative significance: Version type Significance specific point, named functions This tells us that if $g$ is differentiable at a point $x_0$ and $f$ is differentiable at $g(x_0)$, then $f \circ g$ is differentiable at $x_0$. generic point, named functions, point notation If $g$ is a differentiable function and $f$ is a differentiable function on the intersection of its domain with the range of $g$, then $f \circ g$ is a differentiable function. generic point, named functions, point-free notation We can deduce properties of $(f \circ g)'$ based on properties of $f',g',f,g$. In particular, if $f'$ and $g'$ are both continuous functions, so is $(f \circ g)'$. Another way of putting this is that if $f$ and $g$ are both continuously differentiable functions, so is $f \circ g$. ### Computational feasibility significance Each of the versions has its own computational feasibility significance: Version type Significance specific point, named functions If we know the values (in the sense of numerical values) $g'(x_0)$ and $f'(g(x_0))$, we can use these to compute $(f \circ g)'(x_0)$. generic point, named functions This tells us that knowledge of the general expressions for the derivatives of $f$ and $g$ (along with expressions for the functions themselves) allows us to compute the general expression for the derivative of $f \circ g$. ### Computational results significance Shorthand Significance significance of derivative being zero If $\! g'(x_0) = 0$, and $f$ is differentiable at $g(x_0)$, then $\! (f \circ g)'(x_0) = 0$. Note that the conclusion need not follow if $f$ is not differentiable at $g(x_0)$. Also, if $\! f'(g(x_0)) = 0$ and $g$ is differentiable at $x_0$, then $(f \circ g)'(x_0) = 0$. Note that it is essential in both cases that the other function be differentiable at the appropriate point. Here are some counterexamples when it's not: [SHOW MORE] significance of sign of derivative The product of the signs of $\! f'(g(x_0))$ and $\! g'(x_0)$ gives the sign of $(f \circ g)'(x_0)$. In particular, if both have the same sign, then $(f \circ g)'(x_0)$ is positive. If both have opposite signs, then $(f \circ g)'(x_0)$ is negative. This is related to the idea that a composite of increasing functions is increasing, and similar ideas. significance of uniform bounds on derivatives If $\! f'$ and $\! g'$ are uniformly bounded, then so is $(f \circ g)'$, with a possible uniform bound being the product of the uniform bounds for $\! f'$ and $\! g'$. ## Compatibility checks ### Associative symmetry This is a compatibility check for showing that for a composite of three functions $f_1 \circ f_2 \circ f_3$, the formula for the derivative obtained using the chain rule is the same whether we associate it as $f_1 \circ (f_2 \circ f_3)$ or as $(f_1 \circ f_2) \circ f_3$. • Derivative as $f_1 \circ (f_2 \circ f_3)$. We first apply the chain rule for the pair of functions $(f_1, f_2 \circ f_3)$ and then for the pair of functions $(f_2, f_3)$: In point-free notation: $(f_1 \circ (f_2 \circ f_3))' = (f_1' \circ (f_2 \circ f_3)) \cdot (f_2 \circ f_3)' = (f_1' \circ (f_2 \circ f_3)) \cdot (f_2' \circ f_3) \cdot f_3'$ In point notation (i.e., including a symbol for the point where the function is applied): $(f_1 \circ (f_2 \circ f_3))'(x) = f_1'(f_2 \circ f_3(x))(f_2 \circ f_3)'(x) = f_1'(f_2(f_3(x)))(f_2 \circ f_3)'(x) = f_1'(f_2(f_3(x)))f_2'(f_3(x))f_3'(x)$ • Derivative as $(f_1 \circ f_2) \circ f_3$. We first apply the chain rule for the pair of functions $(f_1 \circ f_2, f_3)$ and then for the pair of functions $(f_1, f_2)$: In point-free notation: $((f_1 \circ f_2) \circ f_3)' = ((f_1 \circ f_2)' \circ f_3) \cdot f_3' = ((f_1' \circ f_2) \cdot f_2') \circ f_3) \cdot f_3' = ((f_1' \circ f_2) \circ f_3) \cdot (f_2' \circ f_3) \cdot f_3'$ In point notation (i.e., including a symbol for the point where the function is applied): $((f_1 \circ f_2) \circ f_3)'(x) = ((f_1 \circ f_2)' \circ f_3)(x)f_3'(x) = (f_1 \circ f_2)'(f_3(x))f_3'(x) = f_1'(f_2(f_3(x)))f_2'(f_3(x))f_3'(x)$ ### Compatibility with linearity Consider functions $f_1,f_2,g$. We have that: $(f_1 + f_2) \circ g = (f_1 \circ g) + (f_2 \circ g)$ The function $(f_1 + f_2) \circ g$ can be differentiated either by differentiating the left side or by differentiating the right side. The compatibility check is to ensure that we get the same result from both methods: • Left side: In point-free notation: $\! ((f_1 + f_2) \circ g)' = ((f_1 + f_2)' \circ g) \cdot g' = ((f_1' + f_2') \circ g) \cdot g' = ((f_1' \circ g) + (f_2' \circ g)) \cdot g' = ((f_1' \circ g) \cdot g') + ((f_2' \circ g) \cdot g')$ In point notation (i.e., including a symbol for the point of application): $\! ((f_1 + f_2) \circ g)'(x) = (f_1 + f_2)'(g(x))g'(x) = (f_1'(g(x)) + f_2'(g(x)))g'(x) = f_1'(g(x))g'(x) + f_2'(g(x))g'(x)$ • Right side: In point-free notation: We get $\! (f_1 \circ g + f_2 \circ g)' = (f_1 \circ g)' + (f_2 \circ g)' = ((f_1' \circ g) \cdot g') + ((f_2' \circ g) \cdot g')$. In point notation: $(f_1 \circ g + f_2 \circ g)'(x) = (f_1 \circ g)'(x) + (f_2 \circ g)'(x) = f_1'(g(x))g'(x) + f_2'(g(x))g'(x)$ Thus, we get the same result on both sides, indicating compatibility. Note that it is not in general true that $f \circ (g_1 + g_2) = (f \circ g_1) + (f \circ g_2)$, so there is no compatibility check to be made there. ### Compatibility with product rule Consider functions $f_1,f_2,g$. We have that: $(f_1 \cdot f_2) \circ g = (f_1 \circ g) \cdot (f_2 \circ g)$ The function $(f_1 \cdot f_2) \circ g$ can be differentiated either by differentiating the left side or by differentiating the right side. The two processes use the product rule for differentiation in different ways. The compatibility check is to ensure that we get the same result from both methods: • Left side: In point-free notation: $\! ((f_1 \cdot f_2) \circ g)' = ((f_1 \cdot f_2)' \circ g) \cdot g' = ((f_1' \cdot f_2 + f_1 \cdot f_2') \circ g) \cdot g' = ((f_1' \cdot f_2) \circ g) \cdot g' + ((f_1 \cdot f_2') \circ g) \cdot g'$ In point notation: $\! ((f_1 \cdot f_2) \circ g)' = ((f_1 \cdot f_2)'(g(x)) g'(x) = (f_1'(g(x))f_2(g(x)) + f_1(g(x))f_2'(g(x))) g'(x)$ • Right side: In point-free notation: $\! ((f_1 \circ g) \cdot (f_2 \circ g))' = (f_1 \circ g)' \cdot (f_2 \circ g) + (f_1 \circ g) \cdot (f_2 \circ g)' = (f_1' \circ g) \cdot g' \cdot (f_2 \circ g) + (f_1 \circ g) \cdot (f_2' \circ g) \cdot g'$ $\! = [(f_1' \circ g) \cdot (f_2 \circ g)] \cdot g' + [(f_1 \circ g) \cdot (f_2' \circ g)] \cdot g' = ((f_1' \cdot f_2) \circ g) \cdot g' + ((f_1 \cdot f_2') \circ g) \cdot g'$ In point notation: $\! ((f_1 \circ g) \cdot (f_2 \circ g))'(x) = (f_1 \circ g)'(x)(f_2 \circ g)(x) + (f_1 \circ g)(x)(f_2 \circ g)'(x) = (f_1'(g(x))g'(x)f_2(g(x)) + f_1(g(x))g'(x)f_2'(g(x))$ $\! = (f_1'(g(x))f_2(g(x)) + f_1(g(x))f_2'(g(x))) g'(x)$ Note that it is not in general true that $f \circ (g_1 \cdot g_2) = (f \circ g_1) \cdot (f \circ g_2)$, so no compatibility check needs to be made there. ### Compatibility with notions of order This section explains why the chain rule is compatible with notions of order $\operatorname{ord}$ that satisfy: • $\operatorname{ord}(f') = \operatorname{ord}(f) - 1$ • $\operatorname{ord}(f \circ g) = \operatorname{ord}(f)\operatorname{ord}(g)$ • $\operatorname{ord}(f \cdot g) = \operatorname{ord}(f) + \operatorname{ord}(g)$ Suppose $\operatorname{ord}(f) = m$ and $\operatorname{ord}(g) = n$. Then we have the following: • $(f \circ g)'$ has order $mn -1$: First, note that $f \circ g$ has order $mn$ by the product relation for order. Next, note that differentiating pushes the order down by one. • $(f' \circ g) \cdot g'$ has order $mn - 1$: Note that $f' \circ g$ has order $(m - 1)n$ and $g'$ has order $n - 1$. Adding, we get <math(m - 1)n + n - 1 = mn - 1$.

Some examples of the notion of order which illustrate this are:

• For nonzero polynomials, the order notion above can be taken as the degree of the polynomial.
• For functions that are zero at a particular point, the order notion above can be taken as the order of zero at the point. Note that in this case, the order of zero for $f$ will be calculated at 0 rather than the original point at which $g$ is evaluated.

## Examples

### Sanity checks

We first consider examples where the chain rule for differentiation confirms something we already knew by other means:

Case on $f$ Case on $g$ $(f \circ g)'$ Direct justification, without using the chain rule Justification using the chain rule, i.e., by computing $(f' \circ g) \cdot g'$
a constant function any differentiable function zero function $f \circ g$ is a constant function, so its derivative is the zero function. By the chain rule, $(f \circ g)'(x) = f'(g(x))g'(x)$. $f$ being constant forces $f'(g(x))$ to be zero everywhere, hence the product $f'(g(x))g'(x)$ is also zero everywhere. Thus, $(f \circ g)'$ is also zero everywhere.
any differentiable function a constant function with value $k$ zero function $f \circ g$ is a constant function with value $f(k)$, so its derivative is the zero function. By the chain rule, $(f \circ g)'(x) = f'(g(x))g'(x)$. $g$ being constant forces that $g'(x) = 0$ everywhere, hence the product $f'(g(x))g'(x)$ is also zero everywhere. Thus, $(f \circ g)'$ is also zero everywhere.
the identity function, i.e., the function $x \mapsto x$ any differentiable function $\! g'$ $f \circ g = g$, so $(f \circ g)' = g'$. $(f \circ g)' = (f' \circ g) \cdot g'$. Since $f$ is the function $x \mapsto x$, its derivative is the function $x \mapsto 1$. Plugging this in, we get that $f' \circ g$ is also the constant function $x \mapsto 1$, so $(f \circ g)' = 1g' = g'$.
any differentiable function the identity function $\! f'$ $f \circ g = f$, so $(f \circ g)' = f'$. $(f \circ g)' = (f' \circ g) \cdot g'$. Since $g$ is the identity function, $g'$ is the function $x \mapsto 1$. Also, $f' \circ g = f'$. Thus, $(f \circ g)' = f' \cdot 1 = f'$.
the square function any differentiable function $\! x \mapsto 2g(x)g'(x)$ $f(g(x)) = (g(x))^2$ and hence its derivative can be computed using the product rule for differentiation. It comes out as $2g(x)g'(x)$. $(f \circ g)' = (f' \circ g) \cdot g'$. $f'$ is the derivative of the square function, and therefore is $x \mapsto 2x$. Thus, $\! f'(g(x)) = 2g(x)$. We thus get $(f \circ g)' = 2g(x)g'(x)$.
a one-one differentiable function the inverse function of $f$ 1 $f(g(x)) = x$ for all $x$, so the derivative is the function 1. $(f \circ g)' = (f' \circ g) \cdot g'$. By the inverse function theorem, we know that $g' = 1/(f' \circ g)$, so plugging in, we get $(f \circ g)' = (f' \circ g) \cdot 1/(f' \circ g) = 1$.

### Nontrivial examples

Here are some examples that cannot be computed using methods other than the chain rule:

Consider the sine of square function:

$x \mapsto \sin(x^2)$.

We use the chain rule for differentiation viewing the function as the composite of the square function on the inside and the sine function on the outside:

$\frac{d}{dx}[\sin(x^2)] = \frac{d(\sin(x^2))}{d(x^2)} \frac{d(x^2)}{dx} = (\cos(x^2))(2x) = 2x\cos(x^2)$