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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

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.

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) = $7·e^x ⇒ f’(x) = 7·\frac{d}{dx}e^x = 7·e^x.$

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·xn-1 ⇨ f’(x) =[Product rule and induction hypothesis for n-1] (xn-1)·1+(n-1)xn-2·x= xn-1 + (n-1)xn-1 = n·xn-1

If f(x)=xn, n ≥ 0, then f’(x) = n·xn-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)=nxn-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) = $\sqrt[5]{x^3} ⇒ f’(x) = \frac{3}{5}·x^{\frac{3}{5}-1} = \frac{3}{5}·x^{\frac{-2}{5}} = \frac{3}{5·\sqrt[5]{x^2}}$

(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) = ex + ln(x) ⇒ f’(x) = (ex)’ + (ln(x))’ = $e^x+\frac{1}{x}.$

(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·ex ⇒ f’(x) = ex + xex.
  3. f(x) = x2·cos(x) ⇒ f’(x) = 2xcos(x)-x2sin(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) = $\frac{(x^2-1)^3}{(x^2+1)} ⇒ f’(x) = \frac{3(x^2-1)^2·2x·(x^2+1)-2x(x^2-1)^3}{(x^2+1)^2} = \frac{6x(x^2-1)^2·(x^2+1)-2x(x^2-1)^3}{(x^2+1)^2} = \frac{2x(x^2-1)^2(3(x^2+1)-(x^2-1))}{(x^2+1)^2} = \frac{2x(x^2-1)^2(3x^2+3-x^2+1)}{(x^2+1)^2} = \frac{2x(x^2-1)^2(2x^2+4)}{(x^2+1)^2} = \frac{2x(x^2-1)^2·2(x^2+2)}{(x^2+1)^2} = \frac{4x·(x^2-1)^2·(x^2+2)}{(x^2+1)^2}$

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

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
  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.
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