Chain Rule

Chain Rule#

When we have one function inside another function (this is called a composite function), we differentiate using the chain rule. An example of a composite function is $y=\sin(x^2 + 1)$. There are two functions $\sin(\ldots)$ and $x^2+1$. We are applying the $\sin$ function to $x^2+1$, thus making it a function inside a function, or a composite function.

The general form of a composite function is:

\[ y = f[g(x)], \]

where $f$ and $g$ are both functions. In the above example $y=\sin(x^2+1)$, we would have $f[g(x)] = \sin[g(x)]$ and $g(x) = x^2+1$.

To use the chain rule, we have the following steps:

Introduce a new variable $u$ to be equal to $g(x)$.
Substitute $u=g(x)$ into the expression $y=f[g(x)]$, so that $y=f(u)$.
Find $\frac{dy}{du}$ and $\frac{du}{dx}$.
The derivative $\frac{dy}{dx}$ can be found by this equation

\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]

and then subtitute back into $u=g(x)$.

The chain rule can be summerised as:

\[ \text{If }y=f[g(x)]\text{ let }u=g(x)\text{ hence }y=f[g(x)] = f(u)\text{ then }\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx} \]

Below, we follow the steps of the chain rule to differentiate $y = sin(x^2 + 1)$.

Introduce a new variable $u$ to be equal to $g(x)$. For this example, $u = x^2+1$.
Substitute $u=g(x)$ into the expression $y=f[g(x)]$, so that $y=f(u)$. As $u=x^2+1$, this would make $y = \sin(u)$.
Find $\frac{dy}{du}$ and $\frac{du}{dx}$. In this example, we have:

\[ y = \sin(u) \Rightarrow \frac{dy}{du} = \cos(u) \]

and

\[ u = x^2 + 1 \Rightarrow \frac{du}{dx} = 2x \]
The derivative $\frac{dy}{dx}$ can be found by the equation $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ and then substitute back in $u=g(x)$.

\[ \frac{dy}{dx} = \cos(u)\times 2x = 2x\cos(x^2+1) \]

Python and sympy is able to use the chain rule.

from sympy import symbols, diff, sin

x = symbols('x')

diff(sin(x ** 2 + 1))

\[\displaystyle 2 x \cos{\left(x^{2} + 1 \right)}\]

Example

Differentiate $y=(x^2 + 2)^3$ using the chain rule.

Solution: We take $u = x^2 + 2$ and therefore $y=u^3$.

Then we differentiate each of those to find:

\[ \frac{du}{dx} = 2x \;\;\;\;\; \frac{dy}{du} = 3u^2 \]

From the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}
Substitute in what we know for $\frac{dy}{du}$ and $\frac{du}{dx}$: $\frac{dy}{dx} = 3u^2 \times 2x$
Substitute that $u$ is $x^2+2$: $\frac{dy}{dx} = 6x(x^2+2)^2$.

Example

Given $y=ln(x^3)$, find $\frac{dy}{dx}$.

Solution: We take $u=x^3$, which menas that $y = \ln(u)$.

Then differentiate to find that:

\[ \frac{du}{dx} = 3x^2 \;\;\;\;\; \frac{dy}{du} = \frac{1}{u} \]

Find the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$
We can see that: $\frac{dy}{dx} = \frac{1}{u} \times 3x^2$
Substitute that $u$ is $x^3$: $\frac{dy}{dx} = \frac{1}{x^3} \times 3x^2$
Simplify further: $\frac{dy}{Dx} = \frac{3}{x}$

Example

Differentiate $y=\cos(e^{2x})$

Solution: Let $u=e^{2x}$, therefore, $y=\cos(u)$.

Differentiate to find:

\[ \frac{du}{dx} = 2e^{2x} \;\;\;\;\; \frac{dy}{du} = -\sin(u) \]

Using the chain rule: $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$
We can show that: $\frac{dy}{dx} = -\sin(u) \times 2e^{2x}
Substitute that $u=e^{2x}$
Simplify further: $\frac{dy}{dx} = -2e^{2x}\sin(e^{2x})$.

We can compute the three examples above with Python as follows:

diff((x ** 2 + 2) ** 3)

\[\displaystyle 6 x \left(x^{2} + 2\right)^{2}\]

from sympy import log

diff(log(x ** 3))

\[\displaystyle \frac{3}{x}\]

from sympy import cos, exp

diff(cos(exp(2 * x)))

\[\displaystyle - 2 e^{2 x} \sin{\left(e^{2 x} \right)}\]

Maxwell Boltzmann Distribution

The Maxwell Boltzmann distribution is a probability distribution of finding particles at certains speed $v$ in 3-dimensions. It has the form of:

\[ f(x) = Av^2e^{-Bv^2}, \]

where $A$ and $B$ are positive constants. Using the chain rule and the product rule, find $\frac{d}{dv}f(v)$.

Solution: This example involves two steps, first of all the product rule is required to differentiate the whole thing as $f(v)$ is a product of two functions. The chain rule is also needed to differentiate the exponential. Let’s call $P = Av^2$ and $Q = e^{-Bv^2}$, thus making $f(v) = P\times Q$. From the product rule:

\[ \frac{d}{dv}f(v) = P\frac{dQ}{dv} + Q\frac{dP}{dv} \]

We have $\frac{dP}{dv} = 2Av$, but we will need to use the chain rule to find $\frac{dQ}{dv}.

We let $u=-Bv^2$, making $Q = e^u$. We can then calculate that $\frac{du}{dv} = -2Bv$ and that $\frac{dQ}{du} = e^u$.

From here, the chain rule tells us that: $\frac{dQ}{dv} = \frac{dQ}{du} \times \frac{du}{dv}$
We put in what we have for $\frac{dQ}{du}$ and $\frac{du}{dv}$: $\frac{dQ}{dv} = e^u \times (-2Bv)$
Substitute $u = -Bv^2$: $\frac{dQ}{dv} = -2Bve^{-Bv^2}$

This can then be substituted into the previous equation.

This gives us: $\frac{d}{dv}f(v) = P\frac{dQ}{dv} + Q\frac{dP}{dv}$
Substitute in what we have found: $\frac{d}{dv}f(v) = Av^2(-2Bv\times e^{Bv^2}) + e^{-Bv^2} \times 2Av$
This can be simplified: $\frac{d}{dv}f(v) = -2 ABv^3e^{-Bv^2} + 2Ave^{-Bv^2}$
Take a factor of $2Ave^{-Bv2}$ out: $\frac{d}{dv}f(v) = 2Ave^{-Bv^2}(1-Bv^2)$

With Python, this rather complex example is simple to get the solution to.

from sympy import simplify

A, B, v = symbols('A B v')

simplify(diff(A * v ** 2 * exp(-B * v ** 2), v))

\[\displaystyle 2 A v \left(- B v^{2} + 1\right) e^{- B v^{2}}\]