5.2 Linearity of Differentiation, Product Rule, And Quotient Rule

1. Basic Derivatives

The derivative of a function $f(x)$ is a new function denoted by $f'(x)$. As explained in the preceding section, $f'(x)$ indicates the rate of change, or otherwise the gradient of $f(x)$ at any particular point $x$.

We have seen that the derivative of the function $f(x) = x^{2}$ is $f'(x) = 2x$. We can similarly show that for $f(x) = x^{3}$ the derivative is $f'(x) = 3x^{2}$. In general, the derivative of the power function $f(x) = x^{n}$ is:

$f'(x) = nx^{n-1}$

$f(x) = x^n$	$f'(x) = nx^{n-1}$
$x^{10}$	$10x^9$
$x^{4}$	$4x^3$
$x^{3}$	$3x^2$
$x^{2}$	$2x$
$x$	$1$

$f(x) = x^n$	$f'(x) = nx^{n-1}$
$x^{-10}$	$-10x^{-11}$
$x^{-3}$	$-3x^{-4}$
$x^{-2}$	$-2x^{-3}$
$x^{-1}$	$-x^{-2}$

$f(x) = x^n$	$f'(x) = nx^{n-1}$
$x^{6.4}$	$6.4x^{5.4}$
$x^{3/2}$	$\dfrac{3}{2}x^{1/2}$
$x^{5/3}$	$\dfrac{5}{3}x^{2/3}$
$x^{1/2}$	$\dfrac{1}{2}x^{-1/2}$

Let us especially elaborate on the power formula $(x^n)' = nx^{n-1}$. This formula applies strictly for negative values and rational values of $n$ as well!

Derivatives of Most Common Functions

$f(x)$	$f'(x)$
$x^n$	$nx^{n-1}$
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$e^x$	$e^x$
$\ln x$	$\dfrac{1}{x}$
$\sqrt{x}$	$\dfrac{1}{2\sqrt{x}}$
$c$ (constant)	$0$

EXAMPLE 1

Show that (a) $\left(\dfrac{1}{x^2}\right)' = \dfrac{-2}{x^3}$ and (b) $(\sqrt{x})' = \dfrac{1}{2\sqrt{x}}$

Solution:

(a) Since $\dfrac{1}{x^2} = x^{-2}$, the derivative evaluates strictly to $-2x^{-3} = \mathbf{\dfrac{-2}{x^3}}$.
(b) Since $\sqrt{x} = x^{1/2}$, the derivative evaluates strictly to $\dfrac{1}{2}x^{-1/2} = \dfrac{1}{2x^{1/2}} = \mathbf{\dfrac{1}{2\sqrt{x}}}$.

EXAMPLE 2

Let $f(x) = x^7$. Find the following values:

(a) $f(0)$, $f(1)$, and $f(2)$
(b) $f'(x)$
(c) $f'(0)$, $f'(1)$, and $f'(2)$
(d) the rate of change of $f(x)$ at $x=2$
(e) the gradient of $f(x)$ at $x=2$

Solution:

(a) $f(0) = 0$, $f(1) = 1$, and $f(2) = 128$
(b) $f'(x) = 7x^6$
(c) $f'(0) = 0$, $f'(1) = 7 \cdot 1^6 = 7$, and $f'(2) = 7 \cdot 2^6 = 448$
(d) It evaluates exactly to $f'(2) = \mathbf{448}$
(e) It evaluates identically to $f'(2) = \mathbf{448}$

2. Notation

If $y = f(x)$, the derivative is commonly denoted by the following structural symbols:

$y'$ or $f'(x)$ or $\dfrac{dy}{dx}$ or $\dfrac{d}{dx}f(x)$

The specific derivative evaluated at a static point, say $x=2$, is denoted mathematically by:
$f'(2)$ or $\left. \dfrac{dy}{dx} \right|_{x=2}$

For example, if $y = f(x) = x^3$, we can cleanly write:
$(x^3)' = 3x^2$ or $\dfrac{dy}{dx} = 3x^2$ or $y' = 3x^2$ or $\dfrac{d}{dx}x^3 = 3x^2$
Moreover, $f'(2) = 12$ or $\left. \dfrac{dy}{dx} \right|_{x=2} = 12$.

NOTICE:

The procedural operation of finding the derivative is formally called differentiation.
The GDC does fundamentally provide $f'(x)$, but strictly evaluates the numerical derivative exactly at a specific given point (e.g. $f'(2)$).
The fractional notation $\dfrac{dy}{dx}$ is extremely indicative when utilizing isolated operational variables. For instance, if $s = t^2$, then $\dfrac{ds}{dt} = 2t$. If $P = \ln Q$, then $\dfrac{dP}{dQ} = \dfrac{1}{Q}$. Since $Q = e^P$, we logically establish $\dfrac{dQ}{dP} = e^P$.

3. Rules of Differentiation

Rule (1): Addition and Subtraction

$(f+g)' = f' + g'$ and $(f-g)' = f' - g'$

EXAMPLE 3

For $f(x) = x^5 + x^3$, the derivative evaluates to $f'(x) = \mathbf{5x^4 + 3x^2}$
For $g(x) = x^7 - e^x + \sin x - x + 5$, the derivative evaluates to $$ g'(x) = \mathbf{7x^6 - e^x + \cos x - 1} $$

Rule (2): Scalar Multiplication

$(af)' = a \cdot f'$

(where $a$ is a constant real number)

EXAMPLE 4

For $f(x) = 3\sin x$, the derivative evaluates to $f'(x) = \mathbf{3\cos x}$
For $g(x) = 7e^x$, the derivative evaluates to $g'(x) = \mathbf{7e^x}$
For $h(x) = 5x^3$, the derivative evaluates to $h'(x) = 5(3x^2) = \mathbf{15x^2}$

Let us dynamically combine Rule (1) and Rule (2): $(af + bg)' = af' + bg'$

EXAMPLE 5

For $f(x) = 2x^3 - 3x^2 + 7x + 5$, the derivative evaluates to $f'(x) = \mathbf{6x^2 - 6x + 7}$
For $g(x) = 5x^7 + 3\ln x - 7\cos x$, the derivative evaluates to $$ g'(x) = \mathbf{35x^6 + \dfrac{3}{x} + 7\sin x} $$

NOTICE: Alternative Differential Notation

The differentiation rules above map equivalently under explicit differential operators:

$\dfrac{d}{dx}[f(x) \pm g(x)] = \dfrac{d}{dx}f(x) \pm \dfrac{d}{dx}g(x)$
$\dfrac{d}{dx}[af(x)] = a\dfrac{d}{dx}f(x)$
$\dfrac{d}{dx}[af(x) + bg(x)] = a\dfrac{d}{dx}f(x) + b\dfrac{d}{dx}g(x)$

Rule (3): The Product Rule

$(f \cdot g)' = f' \cdot g + f \cdot g'$

Be Careful!!! If $f(x) = x^5\sin x$, then $f'(x)$ is NOT $(5x^4)(\cos x)$. We must definitively follow the strict product rule sequence above.

EXAMPLE 6

For $f(x) = x^5\sin x$, $$ f'(x) = (x^5)'\sin x + x^5(\sin x)' = \mathbf{5x^4\sin x + x^5\cos x} $$
For $f(x) = x\ln x$, $$ f'(x) = (x)'\ln x + x(\ln x)' = 1\cdot\ln x + x\left(\dfrac{1}{x}\right) = \mathbf{\ln x + 1} $$
For $g(x) = x^2e^x$, $$ g'(x) = \mathbf{2xe^x + x^2e^x} $$
For $h(x) = 2x^3\cos x$, $$ h'(x) = \mathbf{6x^2\cos x - 2x^3\sin x} $$

Rule (4): The Quotient Rule

$\left(\dfrac{f}{g}\right)' = \dfrac{f' \cdot g - f \cdot g'}{g^2}$

EXAMPLE 7

For $f(x) = \dfrac{x^3}{\sin x}$, $$ f'(x) = \dfrac{(x^3)'\sin x - x^3(\sin x)'}{(\sin x)^2} = \mathbf{\dfrac{3x^2\sin x - x^3\cos x}{\sin^2 x}} $$
For $g(x) = \dfrac{3x+5}{4x+1}$, $$ g'(x) = \dfrac{3(4x+1) - 4(3x+5)}{(4x+1)^2} = \dfrac{12x + 3 - 12x - 20}{(4x+1)^2} = \mathbf{\dfrac{-17}{(4x+1)^2}} $$
For $h(x) = \dfrac{x^3-5x}{x^2+1}$, $$ \begin{aligned} h'(x) &= \dfrac{(3x^2-5)(x^2+1) - (x^3-5x)(2x)}{(x^2+1)^2} = \dfrac{3x^4 + 3x^2 - 5x^2 - 5 - 2x^4 + 10x^2}{(x^2+1)^2} \\ &= \mathbf{\dfrac{x^4 + 8x^2 - 5}{(x^2+1)^2}} \end{aligned} $$

EXAMPLE 8

Sometimes, we can bypass the absolute quotient rule by pre-modifying the rational expression. Evaluate $f'(x)$ for $f(x) = \dfrac{x^3-2x+1}{x}$

Method A (Standard Quotient Rule): $$ \begin{aligned} f'(x) &= \dfrac{(3x^2-2)x - (x^3-2x+1)(1)}{x^2} = \dfrac{3x^3 - 2x - x^3 + 2x - 1}{x^2} \\ &= \dfrac{2x^3 - 1}{x^2} = \mathbf{2x - \dfrac{1}{x^2}} \end{aligned} $$

Method B (Algebraic Split):
We modify $f(x)$ by splitting the independent numerator fractions $$ f(x) = \dfrac{x^3}{x} - \dfrac{2x}{x} + \dfrac{1}{x} = x^2 - 2 + x^{-1} $$ Differentiating term-by-term yields $$ f'(x) = 2x - 0 - x^{-2} = \mathbf{2x - \dfrac{1}{x^2}} $$

4. Higher Derivatives

We can mathematically continue differentiating the 1st derivative $f'(x)$ to isolate and find the 2nd derivative $f''(x)$, the 3rd derivative $f'''(x)$, and recursively so on.
The nth derivative is classically denoted by formal index $f^{(n)}(x)$.

Alternative standard notation:

$f''(x)$ can identically be written as $\dfrac{d^2 y}{dx^2}$ or $\dfrac{d^2}{dx^2}f(x)$

$f'''(x)$ can identically be written as $\dfrac{d^3 y}{dx^3}$ or $\dfrac{d^3}{dx^3}f(x)$

EXAMPLE 9

For $f(x) = x^5$, $$ f'(x) = 5x^4 \quad \implies \quad f''(x) = 20x^3 \quad \implies \quad f'''(x) = 60x^2 $$
For $g(x) = \sin x$, $$ g'(x) = \cos x \quad \implies \quad g''(x) = -\sin x \quad \implies \quad g'''(x) = -\cos x $$
For $h(x) = e^x$, $$ h'(x) = e^x \quad \implies \quad h''(x) = e^x \quad \implies \quad h'''(x) = e^x $$ Clearly $h^{(4)}(x) = e^x$ and naturally in general $h^{(n)}(x) = e^x$.