General Differentiation rules

The saddest aspect of life right now is that science gathers knowledge faster than society gathers wisdom, Isaac Asimov.

Recall

The derivative of a function at a specific input value describes the instantaneous rate of change. Geometrically, when the derivative exists, it is the slope of the tangent line to the graph of the function at that point. Algebraically, it represents the ratio of the change in the dependent variable to the change in the independent variable as the change approaches zero.

Definition. A function f(x) is differentiable at a point “a” in its domain, if its domain contains an open interval around “a”, and the following limit exists $f'(a) = L = \lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$. More formally, for every positive real number ε, there exists a positive real number δ, such that for every h satisfying 0 < |h| < δ, then the following inequality holds |L-$\frac {f(a+h)-f(a)}{h}$|< ε.

 

General Differentiation rules

Differentiation rules allow us to compute derivatives efficiently without returning to the limit definition each time. All the rules below can be justified rigorously from the definition of the derivative.

• Constant Rule. Let f(x) = c, where c is a constant, then f'(x) = 0.

f’(x) = $\lim_{h \to 0} \frac{f(x+h)-f(h)}{h} = \lim_{h \to 0} \frac{c-c}{h} = \lim_{h \to 0} 0 = 0.$

Examples

1. f(x) = 5 ⇒ f’(x) = 0.
2. f(x) = -3 ⇒ f’(x) = 0.

• Constant Multiple Rule. The derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function. Let c be a constant and g(x) a differentiable function, then the function f(x) = c·g(x) is also differentiable and f'(x) = (c·g(x))' = c·g'(x).

f’(x) = $\lim_{h \to 0} \frac{f(x+h)-f(h)}{h} = \lim_{h \to 0} \frac{c·g(x+h)-c·g(h)}{h} = \lim_{h \to 0} c·\frac{g(x+h)-g(h)}{h}$[Constant multiple law for limits] $c·\lim_{h \to 0} \frac{g(x+h)-g(h)}{h}$ =[g(x) is a differentiable function] c·g’(x).

Examples

1. f(x) = 3x² ⇒ $f'(x) = 3·\frac{d}{dx}x^2 = 3·(2x) = 6x.$
2. f(x) = -4·sin(x) ⇒ $f'(x) = -4·\frac{d}{dx}sin(x) = -4·cos(x).$
3. f(x) = $7·e^x ⇒ f'(x) = 7·\frac{d}{dx}e^x = 7·e^x.$

• Power rule. If f(x) = xⁿ, where $n \in \mathbb{Z}$ then f’(x) = nx^n-1. Proof by induction:

1. n = 0, f(x) = 1. $\lim_{h \to 0}\frac{1 -1}{h} = 0$, n = 1, f(x) = x. $\lim_{h \to 0}\frac{x+h-x}{h} = \lim_{h \to 0}\frac{h}{h} = \lim_{h \to 0} 1 = 1 \implies f'(x) = 1 = 1·x^0$.
2. Inductive step: Assume the rule holds for n-1. f(x)=xⁿ = x·x^n-1 ⇨ f’(x) =[Product rule and induction hypothesis for n-1] (x^n-1)·1+(n-1)x^n-2·x= x^n-1 + (n-1)x^n-1 = n·x^n-1. If f(x) = xⁿ, n ≥ 0, then f’(x) = n·x^n-1.
3. Extension to negative integers. For m < 0 (m = -n, m > 0), write $x^{m} = \frac{1}{x^{n}}$. Using the Quotient Rule: $ (\frac{1}{x^{n}})' = \frac{u'v - uv'}{v^{2}} = \frac{-nx^{n-1}}{x^{2n}} = -nx^{n-1-2n} = -nx^{-n-1} = mx^{m-1}$

Examples

1. f(x) = x⁴ ⇒ f’(x) = 4·x³.
2. f(x) = $\sqrt{x} ⇒ f'(x) = \frac{1}{2}·x^{\frac{1}{2}-1} = \frac{1}{2}·x^{\frac{-1}{2}} = \frac{1}{2·\sqrt{x}}$
3. f(x) = $\sqrt[5]{x^3} ⇒ f'(x) = \frac{3}{5}·x^{\frac{3}{5}-1} = \frac{3}{5}·x^{\frac{-2}{5}} = \frac{3}{5·\sqrt[5]{x^2}}$

• Sum Rule. It states that the derivative of a sum is equal to the sum of the derivatives. Let f(x) and g(x) be differentiable functions. Then, the sum of both functions f(x) + g(x) is also differentiable and (f+g)'(x) = f'(x) + g'(x).

(f+g)’(x) = $\lim_{h \to 0} \frac{(f+g)(x+h)-(f+g)(h)}{h} = \lim_{h \to 0} \frac{f(x+h)+g(x+h)-f(h)-g(h)}{h}$ =[Sum law for limits] $\lim_{h \to 0} \frac{f(x+h)-f(h)}{h} + \lim_{h \to 0} \frac{g(x+h)-g(h)}{h}$ =[f(x) and g(x) are differentiable function’] f’(x) + g’(x).

Examples

1. f(x) = x³ +2x² + 5 ⇒[Power, constant, and sum rule] f’(x) = 3·x² + 4x.
2. f(x) = sin(x) + cos(x) ⇒ f’(x) = cos(x) −sin(x).
3. f(x) = e^x + ln(x) ⇒ f’(x) = (e^x)’ + (ln(x))’ = $e^x+\frac{1}{x}.$

• Product rule: Let f(x) and g(x) be differentiable functions. Then, the product of both functions f(x)·g(x) is also differentiable and (f·g)' = f(x)·g'(x) + g(x)·f'(x).

Start with the definition:

$$ \begin{aligned} (f·g)'(x) &=\lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} \\[2pt] &[\text{Add and subtract f(x+h)g(x) in the numerator: }] \\[2pt] &=\lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)}{h} \\[2pt] &=\lim_{h \to 0} f(x+h)\frac{g(x+h)-g(x)}{h} + g(x)\frac{f(x+h)-f(x)}{h} \\[2pt] &=\lim_{h \to 0}f(x+h)\frac{g(x+h)-g(x)}{h} + \lim_{h \to 0} g(x)\frac{f(x+h)-f(x)}{h} \\[2pt] &=\lim_{h \to 0} f(x+h)·\lim_{h \to 0}\frac{g(x+h)-g(x)}{h}+ \lim_{h \to 0} g(x)·\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\[2pt] &=f(x)·g'(x)+g(x)·f'(x). \end{aligned} $$

Examples

1. f(x) = $(3x^2-1)(x^2+5x+2) ⇒ f'(x) = 6x(x^2+5x+2)+(2x+5)(3x^2-1) = 6x^3+30x^2+12x+6x^3+15x^2-2x-5=12x^{3}+45x^{2}+10x-5.$
2. f(x) = x·e^x ⇒ f’(x) = e^x + xe^x.
3. f(x) = x²·cos(x) ⇒ f’(x) = 2xcos(x)-x²sin(x).

• Quotient rule: Let f(x) and g(x) be differentiable functions. Then, if g(x) ≠ 0, the derivate of the quotient of these functions f/g is also differentiable and could be calculated using the formula, (f/g)'(x) = $\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$.

Start with the definition:

$$ \begin{aligned} (f/g)'(x) &=\lim_{h \to 0} \frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h} \\[2pt] &[\text{Combine the fractions in the numerator}] \\[2pt] &=\lim_{h \to 0} \frac{\frac{f(x+h)g(x)-f(x)g(h+x)}{g(x+h)g(x)}}{h} \\[2pt] &[\text{Add and subtract f(x)g(x) in the numerator}] \\[2pt] &=\lim_{h \to 0} \frac{\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(h+x)}{g(x+h)g(x)}}{h} \\[2pt] &[\text{Split the limit into two terms (using the property that the limit of a difference equals the difference of the limits, provided they exist)}] \\[2pt] &=\lim_{h \to 0} \frac{\frac{g(x)(f(x+h)-f(x))-f(x)(g(h+x)-g(x))}{g(x+h)g(x)}}{h} \\[2pt] &[\text{In the first term, cancel g(x); in the second term, rearrange.}] \\[2pt] &=\lim_{h \to 0} \frac{\frac{g(x)(f(x+h)-f(x))}{g(x+h)g(x)}}{h}-\lim_{h \to 0} \frac{\frac{f(x)(g(h+x)-g(x))}{g(x+h)g(x)}}{h} \\[2pt] &=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}·\lim_{h \to 0} \frac{1}{g(x+h)}-\lim_{h \to 0} \frac{g(x+h)-g(x)}{h}·\lim_{h \to 0} \frac{f(x)}{g(x+h)g(x)} \\[2pt] &\text{Since g is differentiable, it is also continuous} \lim_{h \to 0}g(x+h) = g(x). \text{Because g(x)} \ne 0, \lim_{h \to 0}\frac{1}{g(x+h)} = \frac{1}{g(x)} \\[2pt] &=f'(x)\cdot \frac{1}{g(x)}-g'(x)\cdot \frac{f(x)}{g(x)^2}=\frac{f'(x)}{g(x)}-\frac{f(x)·g'(x)}{g(x)^2} = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2} \end{aligned} $$

1. f(x) = $\frac{1}{x} ⇒ f'(x) = \frac{0·x-1·1}{x^2} = \frac{-1}{x^2}$.
2. f(x) = $\frac{x^2+6}{2x-7}$ ⇒[Apply the quotient rule:] f’(x) = $\frac{2x·(2x-7)-2·(x^2+6)}{(2x-7)^2} = 2\frac{x·(2x-7)-(x^2+6)}{(2x-7)^2} = 2\frac{2x^2-7x-x^2-6}{(2x-7)^2} = \frac{2(x^2-7x-6)}{(2x-7)^2}$.
3. f(x) = tan(x) = $\frac{sin(x)}{cos(x)}$ ⇒ f’(x) =[Quotient Rule] $\frac{cos(x)cos(x)+sin(x)sin(x)}{cos^{2}x} = \frac{1}{cos^{2}x}$
4. f(x) = $\frac{(x^2-1)^3}{(x^2+1)} ⇒ f'(x) = \frac{3(x^2-1)^2·2x·(x^2+1)-2x(x^2-1)^3}{(x^2+1)^2} = \frac{6x(x^2-1)^2·(x^2+1)-2x(x^2-1)^3}{(x^2+1)^2} = \frac{2x(x^2-1)^2(3(x^2+1)-(x^2-1))}{(x^2+1)^2} = \frac{2x(x^2-1)^2(3x^2+3-x^2+1)}{(x^2+1)^2} = \frac{2x(x^2-1)^2(2x^2+4)}{(x^2+1)^2} = \frac{2x(x^2-1)^2·2(x^2+2)}{(x^2+1)^2} = \frac{4x·(x^2-1)^2·(x^2+2)}{(x^2+1)^2}$

• Chain rule. It states that the derivative of f(g(x)) is f'(g(x))⋅g'(x). In words, the derivative of a composite function is the product of the derivative of the outer function evaluated at the inner function (f’(g(x))) and the derivative of the inner function (g’(x)).

Theorem: If $g$ is differentiable at $a$ and $f$ is differentiable at $g(a)$, then $f \circ g$ is differentiable at $a$ and $(f \circ g)'(a) = f'(g(a)) \cdot g'(a)$

Informal Proof (Leibniz Notation):

1. Let $y = f(u)$ and $u = g(t)$.
2. For an infinitesimal $\Delta t \neq 0$, $\Delta u = g(t + \Delta t) - g(t)$ and $\Delta y = f(u + \Delta u) - f(u)$
3. Then, $\frac{\Delta y}{\Delta t} = \frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta t}$
4. Taking the limit as $\Delta t \to 0$ (which implies $\Delta u \to 0$ by continuity of $g$): $\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt}$

Note: $\dfrac{dy}{du}$ is evaluated at $u = g(t)$, i.e., $\dfrac{dy}{du}\bigg|_{u=g(t)}$

Formal Proof:

1. Define an auxiliary function to handle the case when $\Delta u = 0$. Let: $\phi(u) = \begin{cases} \dfrac{f(u) - f(g(a))}{u - g(a)}, & u \neq g(a) \\ f'(g(a)), & u = g(a) \end{cases}$. Since $f$ is differentiable at $g(a)$, $\phi$ is continuous at $g(a)$: $\lim_{u \to g(a)}\phi(u) = \lim_{u \to g(a)} \dfrac{f(u) - f(g(a))}{u - g(a)}= f'(g(a)) = \phi(g(a))$.
2. For all $u$ (including $u = g(a)$): $f(u) - f(g(a)) = \phi(u) \cdot (u - g(a))$
3. Apply to $u = g(x)$, $f(g(x)) - f(g(a)) = \phi(g(x)) \cdot (g(x) - g(a))$ and divide by $x - a$, $\frac{f(g(x)) - f(g(a))}{x - a} = \phi(g(x)) \cdot \frac{g(x) - g(a)}{x - a}$
4. Take the limit as $x \to a$, (i) $g(x) \to g(a)$ (by continuity of $g$); (ii) $\phi(g(x)) \to \phi(g(a)) = f'(g(a))$ (by continuity of $\phi$); (iii) $\dfrac{g(x) - g(a)}{x - a} \to g'(a)$ (g is differentiable at a). Hence, $\lim_{x \to a}\frac{f(g(x)) - f(g(a))}{x - a} = \lim_{x \to a}\phi(g(x)) \cdot \frac{g(x) - g(a)}{x - a} = f'(g(a)) \cdot g'(a)$

Therefore: $(f \circ g)'(a) = f'(g(a)) \cdot g'(a) \quad \blacksquare$

Examples

1. Power of a Trigonometric Function. y = cos²(x), y’= 2·cos(x)·(-sin(x)) = -2cos(x)sin(x) = -sin(2x).
2. Trigonometric Function with Linear Argument. y = sin(nx), y’=cos(nx)·n = n·cos(nx).
3. Square Root of a Polynomial. y = $\sqrt{x^3+x^2-1}, y' = \frac{1}{2\sqrt{x^3+x^2-1}}·(3x^2+2x) = \frac{3x^2+2x}{2\sqrt{x^3+x^2-1}}.$
4. Implicit Differentiation. When y is implicitly defined by $F(x, y) = 0$, treat $y$ as $y(x)$ and apply chain rule: $x^2 + y^2 = 25 \implies 2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$
5. Related Rates. If quantities are related and change with time: $A = \pi r^2 \implies \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}$
6. Logarithmic Differentiation. For $y = f(x)^{g(x)}$, $\ln(y) = g(x)\ln(f(x)) \implies \frac{y'}{y} = g'(x)\ln(f(x)) + g(x)\frac{f'(x)}{f(x)}$
7. Generalized Chain Rule. For $y = f(g(h(x))), \frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$, e.g., $y = \sin(\cos(x^2)), y' = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot 2x = -2x\sin(x^2)\cos(\cos(x^2))$
8. $[f(x)]^n \to[Derivative] n[f(x)]^{n-1} \cdot f'(x)$; $e^{f(x)} \to[Derivative] e^{f(x)} \cdot f'(x)$
9. $\ln(f(x)) \to[Derivative] \dfrac{f'(x)}{f(x)}$; $\sin(f(x)) \to[Derivative] \cos(f(x)) \cdot f'(x)$
10. $\cos(f(x)) \to[Derivative] -\sin(f(x)) \cdot f'(x)$; $\tan(f(x)) \to[Derivative] \sec^2(f(x)) \cdot f'(x)$
11. $a^{f(x)} \to[Derivative] a^{f(x)} \ln(a) \cdot f'(x)$; $\sqrt{f(x)} \to[Derivative] \dfrac{f'(x)}{2\sqrt{f(x)}}$
12. $\arcsin(f(x)) \to[Derivative] \dfrac{f'(x)}{\sqrt{1-[f(x)]^2}}$; $\arctan(f(x)) \to[Derivative] \dfrac{f'(x)}{1+[f(x)]^2}$

• $y = \left(\frac{x+4}{\sqrt{x^2+1}}\right)^3$

1. Let $u = \dfrac{x+4}{\sqrt{x^2+1}}$, so $y = u^3$
2. $\frac{dy}{du} = 3u^2 = 3\left(\frac{x+4}{\sqrt{x^2+1}}\right)^2$
3. $\frac{du}{dx} =[\text{Quotient Rule}] \frac{\sqrt{x^2+1} \cdot 1 - (x+4) \cdot \dfrac{x}{\sqrt{x^2+1}}}{x^2+1} = \frac{\dfrac{(x^2+1) - x(x+4)}{\sqrt{x^2+1}}}{x^2+1} = \frac{x^2 + 1 - x^2 - 4x}{(x^2+1)^{3/2}} = \frac{1-4x}{(x^2+1)^{3/2}}$
4. Apply chain rule: $\frac{dy}{dx} = 3\left(\frac{x+4}{\sqrt{x^2+1}}\right)^2 \cdot \frac{1-4x}{(x^2+1)^{3/2}} = \frac{3(x+4)^2}{x^2+1} \cdot \frac{1-4x}{(x^2+1)^{3/2}} = \frac{3(x+4)^2(1-4x)}{(x^2+1)^{5/2}} = \frac{3(x+4)^2(1-4x)}{\sqrt{(x^2+1)^5}}$

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