Derivative 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 chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. It is the instantaneous rate of change, the ratio of the instantaneous change in the dependent variable to that of the independent variable.

Definition. A function f(x) is differentiable at a point “a” of its domain, if its domain contains an open interval containing “a”, and the limit $\lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$ 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 |L-$\frac {f(a+h)-f(a)}{h}$|< ε.

General rules

• Constant Rule. If f(x) =c, then f'(x) = 0.

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

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 diferenciable function, then 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).

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

• Power rule. If f(x)=xⁿ, then f’(x)=nx^n-1.

n = 0, f(x) = 0. $\lim_{h \to 0}\frac{0 -0}{h} = 0.$

Proof by induction. n = 1, f(x) = x. $\lim_{h \to 0}\frac{x+h-x}{h} = \lim_{h \to 0}\frac{h}{h} = \lim_{h \to 0} 1 = 1.$ ⇨ f’(x) = 1 = 1·x⁰.

Suppose it 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.

Let u(x)=1, v(x)=xⁿ ⇨ [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}$

If f(x)=xⁿ, n=0, ±1, ±2, ±3,… then u’(x)=nx^n-1.

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 the 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).

1. f(x) = x³ +2x² + 5 ⇒[Power, constant, and sum rule] f’(x) = 3·x² + 4x + 5.
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 the functions f(x)·g(x) is also differentiable and (f·g)' = f(x)·g'(x)+g(x)·f'(x).

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

1. f(x) = $(3x^2-1)(x^2+5x+2) ⇒ f’(x) = 6x(x^2+5x+2)+(2x+5)(3x^2-1) = 12x^{3}+45x^{2}+10x-5.$
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 f(x)≠0, the derivate of the quotient of these functions f(x)·g(x) is also differentiable and calculated by the formula, (f/g)'(x) = $\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$.

(f/g)’(x) = $\lim_{h \to 0} \frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h} = \lim_{h \to 0} \frac{\frac{f(x+h)g(x)-f(x)g(h+x)}{g(x+h)g(x)}}{h} = \lim_{h \to 0} \frac{\frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(h+x)}{g(x+h)g(x)}}{h} = \lim_{h \to 0} \frac{\frac{g(x)(f(x+h)-f(x))-f(x)(g(h+x)-g(x))}{g(x+h)g(x)}}{h} = \lim_{h \to 0} \frac{\frac{g(x)(f(x+h)-f(x))}{g(x+h)g(x)}}{h}-\lim_{h \to 0} \frac{\frac{f(x)(g(h+x)-g(x))}{g(x+h)g(x)}}{h} = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}·\lim_{h \to 0} \frac{1}{g(x+h)}-\lim_{h \to 0} \frac{g(x+h)-g(x)}{h}·\lim_{h \to 0} \frac{f(x)}{g(x+h)g(x)} = \frac{f’(x)}{g(x)}-\frac{f(x)·g’(x)}{g^2(x)} = \frac{f’(x)g(x)-f(x)g’(x)}{g^2(x)}$

1. f(x) = $\frac{1}{x} ⇒ f’(x) = \frac{0·x-1·1}{x^2} = \frac{-1}{x^2}$.
2. f(x) = $\frac{x^2+6}{2x-7}$ ⇒ 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)).

Let y = f(x) and x = g(t), then choosing infinitesimal $\Delta t ≠ 0$, we compute the corresponding $\Delta x = g(t +\Delta t)-g(t)$ and $\Delta y = f(x +\Delta t)-f(x)$ ⇒ $\frac{\Delta y}{\Delta t} = \frac{\Delta y}{\Delta x}·\frac{\Delta x}{\Delta t}$. And therefore, in the limit as △t→0, $\frac{dy}{dt} =\frac{dy}{dx}\frac{dx}{dt}$ which is the chain rule. Notice that $\frac{dy}{dx} = \frac{dy}{dx}\bigg|_{g(t)}$

1. y = cos²x, y’= 2·cos(x)·(-sin(x)) = -2cos(x)sin(x).
2. y = sin(nx), y’=cos(nx)·n = n·cos(nx).
3. 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. y = $(\frac{x+4}{\sqrt{x^2+1}})^3 ⇒ f’(x)=3(\frac{x+4}{\sqrt{x^2+1}})^2·\frac{\sqrt{x^2+1}-(x+4)\frac{1}{2}(x^2+1)^{\frac{-1}{2}}(2x)}{x^2+1} = \frac{3(x+4)^2}{x^2+1}·\frac{(x^2+1)^{\frac{-1}{2}}((x^2+1)-x(x+4))}{x^2+1} = \frac{3(x+4)^2(1-4x)}{(x^2+1)^{1+\frac{1}{2}+1}} = \frac{3(x+4)^2(1-4x)}{(x^2+1)^{\frac{5}{2}}} = \frac{3(x+4)^2(1-4x)}{\sqrt{(x^2+1)^5}}$

Summary

1. Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$.
2. Constant Multiple Rule: $\frac{d}{dx}(cf(x)) = c \cdot \frac{d}{dx}(f(x))$.
3. Sum Rule: $\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}(f(x)) + \frac{d}{dx}(g(x))$
4. Product Rule: $\frac{d}{dx}(f(x) \cdot g(x)) = f’(x)g(x) + f(x)g’(x)$.
5. Quotient Rule: $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f’(x)g(x) - f(x)g’(x)}{(g(x))^2}$
6. Chain Rule: $\frac{d}{dx}(f(g(x))) = f’(g(x)) \cdot g’(x)$

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus.
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus, College Algebra and Abstract Algebra.
8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
9. blackpenredpen.