“You are a just a fucking manipulator.” “I see,” I paused, looked down at her, and then I smiled. “I like to think of myself more as a creative mind and an outcome engineer,” Apocalypse, Anawim, #justtothepoint.
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.
- Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$.
- Sum Rule: $\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}(f(x)) + \frac{d}{dx}(g(x))$
- Product Rule: $\frac{d}{dx}(f(x) \cdot g(x)) = f’(x)g(x) + f(x)g’(x)$.
- Quotient Rule: $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f’(x)g(x) - f(x)g’(x)}{(g(x))^2}$
- Chain Rule: $\frac{d}{dx}(f(g(x))) = f’(g(x)) \cdot g’(x)$
- $\frac{d}{dx}(e^x) = e^x, \frac{d}{dx}(\ln(x)) = \frac{1}{x}, \frac{d}{dx}(\sin(x)) = \cos(x), \frac{d}{dx}(\cos(x)) = -\sin(x), \frac{d}{dx}(\tan(x)) = \sec^2(x), \frac{d}{dx}(\arcsin(x)) = \frac{1}{\sqrt{1 - x^2}}, \frac{d}{dx}(\arccos(x)) = -\frac{1}{\sqrt{1 - x^2}}, \frac{d}{dx}(\arctan(x)) = \frac{1}{1 + x^2}.$
Antiderivatives
Antiderivatives are fundamental concepts in calculus. They are the inverse operation of derivatives.
Given a function f(x), an antiderivative, also known as indefinite integral, F, is the function that can be differentiated to obtain the original function, that is, F’ = f, e.g., 3x2 -1 is the antiderivative of x3 -x +7 because $\frac{d}{dx} (x^3-x+7) = 3x^2 -1$. Symbolically, we write F(x) = $\int f(x)dx$.
The process of finding antiderivatives is called integration.
Examples
- $\int sinxdx = -cos(x) + C$ where C is a constant, i.e., the antiderivative is ambiguous up to a constant.
- $\int cosxdx = sin(x) + C$ where C is a constant.
- $\int sec^{2}xdx = tanx + C$ because tan’(x) = sec2(x).
- $\int \frac{dx}{\sqrt{1-x^{2}}}=sin^{-1}x+C$.
- $\int \frac{dx}{1+x^{2}}=tan^{-1}x+C.$
- $\int x^n dx = \frac{1}{n+1}x^{n+1}dx + C$ where n ≠ -1 because $\frac{d}{dx} (\frac{x^{n+1}}{n+1}+C) = x^n$, e.g., $\int (x^2 -3x)dx = \frac{x^3}{3}-\frac{3x^2}{2}+C$
- $\int \frac{x^2+4\sqrt[3]{x}}{x}dx = \int xdx + \int \frac{4x^{\frac{1}{3}}}{x}dx = \frac{x^2}{2} + \int 4x^{\frac{-2}{3}} = \frac{x^2}{2} + 4·\frac{x^{\frac{-2}{3}+1}}{\frac{-2}{3}+1} = \frac{x^2}{2} + 4·\frac{x^{\frac{1}{3}}}{\frac{1}{3}} = \frac{x^2}{2} + 12\sqrt[3]{x} + C.$
- $\int e^{x} dx = e^x + C.$
- More generally, $\int e^{ax} dx = \frac{1}{a}e^{ax}dx + C$ because $\frac{d}{dx} (\frac{e^{ax}}{a}+C) = e^{ax},$ e.g., $\int e^{4x}dx$ = $\frac{1}{4}e^{4x} + C.$
- $\int \frac{dx}{x} = ln|x| + C.$ It is obvious for x positive. Let be x < 0, $\frac{d}{dx}ln|x|=\frac{d}{dx}ln(-x)=\frac{1}{-x}(-1)=\frac{1}{x}$.
- $\int (\frac{x-7}{x})dx = \int 1dx -\int \frac{7}{x}dx = x -7ln|x| + C.$
- $\int x^{3}(x^{4}+2)^{5}dx$ =[Integration by substitution or change of variables, 1. Choose a suitable substitution, e.g., u = x4 + 2. 2. Find the differential of u with respect to x, du = 4x3dx.] 3. Rewrite the integral in terms of the new variable, e,g, $\int u^{5}\frac{du}{4}$. 4. Evaluate the integral in terms of the new variable.
$\int u^{5}\frac{du}{4} = \frac{u^{6}}{24} + C$ =[5. Convert back to the original variable] $\frac{(x^{4}+2)^{6}}{24} + C.$
$\int \frac{1}{u}du = ln|u| + C = ln|lnx| + C$ where C is the constant of integration.
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$\int \frac{x}{1+x^{2}}dx =$ [Integration by substitution or change of variables, u = 1 + x2, du = 2xdx] $\int \frac{1}{2}\frac{du}{u} = \frac{1}{2}ln|u| + C = \frac{1}{2}ln|1+x^2|+C$
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$\int xe^{x^{2}}dx$ =[Integration by substitution or change of variables, u=x2, du = 2xdx] = $\frac{1}{2}\int e^udu = \frac{1}{2}e^u = \frac{1}{2}e^{x^2} + C.$
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$\int xcosxdx$ = [Integration by parts, ∫udv = uv −∫vdu, u = x, dv = cos(x)dx] $xsin(x) -\int sin(x)dx = xsin(x) + cos(x) + C$.
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$\int sinxcosxdx$ =[Integration by substitution or change of variables, u = sinx, du = cosxdx] = $\int udu = \frac{u^2}{2} = \frac{1}{2}sin^2x+C$ = [Using the Pythagorean identity, sin2x = 1 -cos2x] = $\frac{1}{2}(1-cos^2x)+C = \frac{1}{2}-\frac{1}{2}cos^2x+C = -\frac{1}{2}cos^2x+C’$.
Uniqueness of Antiderivatives
Let F and G two different antiderivatives of f(x). If F’ = G’, then F(x) = G(x) + C. They are unique up to a constant.
Proof. If F’ = G’ = f ⇒ (F-G)’ = F’ - G’ = 0 ⇒ F-G is constant, that is, F(x) - G(x) = C ⇒ F(x) = G(x) + C∎
Bibliography
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- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn, and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
- MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
- Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.