18.01: Single Variable Calculus
Section 4
Chain Rule, and Higher Derivatives
Massachusetts Institute of Technology
Last Edit Date: 06/06/2024
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Chain Rule¶
We've got general procedures for differentiating expressions with addition, subtraction, and multiplication. What about composition?
Example 1. $y = f(x) = \sin x$, $x = g(t) = t^2$.
So, $y = f(g(t)) = \sin (t^2)$. To find $\frac{dy}{dt}$, write
| $t_0 = t_0$ | $t = t_0 + \Delta t$ |
| $x_0 = g(t_0)$ | $x = x_0 + \Delta x$ |
| $y_0 = f(x_0)$ | $y = y_0 + \Delta y$ |
$$\frac{\Delta y}{\Delta t} = \frac{\Delta y}{\Delta x} \cdot \frac{\Delta x}{\Delta t}$$
As $\Delta t \rightarrow 0$, $\Delta x \rightarrow 0$ too, because of continuity. So we get:
$$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \leftarrow \text{The chain rule}$$
In the example, $\frac{dx}{dt} = 2t$ and $\frac{dy}{dx} = \cos x$.
So,
$$ \begin{aligned} \frac{d}{dt} \left( \sin (t^2) \right) &= \left( \frac{dy}{dx} \right) \left( \frac{dx}{dt} \right) \\ &= \left( \cos x \right) \left( 2t \right) \\ &= \left( 2t \right) \left( \cos t^2 \right) \\ \end{aligned} $$
Another notation for the chain rule
$$\frac{d}{dt} f(g(t)) = f'(g(t)) g'(t)~~~\left( \text{or $\frac{d}{dx}f(g(x)) = f'(g(x)) g'(x)$}\right)$$
Example 1. (continued) Composition of functions $f(x) = \sin x$ and $g(x) = x^2$
$$(f \circ g)(x) = f(g(x)) = \sin (x^2)$$
$$(g \circ f)(x) = g(f(x)) = \sin^2 (x)$$
Note: $f \circ g \neq g \circ f$ not commutative
Example 2. $\frac{d}{dx} \cos \left( \frac{1}{x} \right) = ?$
Let $u = \frac{1}{x}$
$$ \begin{aligned} \frac{dy}{dx} &= \frac{dy}{du} \frac{du}{dx} \\ \frac{dy}{du} &= - \sin (u);~~~~~\frac{du}{dx} = -\frac{1}{x^2} \\ \frac{dy}{dx} &= \frac{\sin (u)}{x^2} = (- \sin u) \left( \frac{-1}{x^2} \right) = \frac{\sin \left( \frac{1}{x} \right)}{x^2} \end{aligned} $$
Example 3. $\frac{d}{dx} (x^{-n}) = ?$
There are two ways to proceed. $x^{-n} = \left( \frac{1}{x} \right)^n$, or $x^{-n} = \frac{1}{x^n}$
$\frac{d}{dx} (x^{-n}) = \frac{d}{dx} \left( \frac{1}{x^n} \right) = nx^{n - 1} \left( \frac{-1}{x^{2n}} \right) = -nx^{-(n - 1)}x^{-2} = -n x^{-n-1}$
$\frac{d}{dx} (x^{-n}) = \frac{d}{dx} \left( \frac{1}{x^n} \right) = nx^{n - 1} \left( \frac{-1}{x^{2n}} \right) = -nx^{-n-1}~~~\text{(Think of $x^n$ as $u$)}$
Higher Derivatives¶
Higher derivatives are derivatives of derivatives. For instance, if $g = f'$, then $h = g'$ is the second derivative of $f$. We write $h = (f')' = f''$.
Notations¶
| $f'(x)$ | $Df$ | $\frac{df}{dx}$ |
| $f''(x)$ | $D^2f$ | $\frac{d^2f}{dx^2}$ |
| $f'''(x)$ | $D^3f$ | $\frac{d^3f}{dx^3}$ |
| $f^{(n)}(x)$ | $D^{n}f$ | $\frac{d^nf}{dx^n}$ |
Higher derivatives are pretty straightforward - just keep taking the derivative.
Example. $D^nx^n=?$
Start small and look for a pattern.
$$ \begin{aligned} Dx &= 1 \\ D^2x^2 &= D(2x) = 2~~(= 1 \cdot 2)\\ D^3x^3 &= D^2(3x^2) = D(6x) = 6~~(= 1 \cdot 2 \cdot 3)\\ D^4x^4 &= D^3(4x^3) = D^2(12x^2) = D(24x) = 24~~(= 1 \cdot 2 \cdot 3 \cdot 4)\\ D^nx^n &= n!~~\leftarrow~~\text{we guess, based on the pattern we are seeing here.} \end{aligned} $$
The notation $n!$ is called "n factorial" and defined by $n! = n(n - 1) \cdots 2 \cdot 1$
Proof by Induction: We've already checked the base case $n = 1$.
Induction step: Suppose we know $D^nx^n = n!$ ($n^{th}$ case). Show it holds for the $(n + 1)^{st}$ case.
$$ \begin{aligned} D^{n + 1} x^{n + 1} &= D^n (Dx^{x + 1}) = D^n ((n + 1)x^n) = (n + 1)D^n x^n = (n + 1)(n!)\\ D^{n + 1} x^{n + 1} &= (n + 1)! \end{aligned} $$
Proved.