Comprehensive Calculus Notes: Limits, Derivatives, and Integrals

Evaluation of Limits and L'Hôpital's Rule (LH) Applications: * a) For the limit $\lim_{x \to 2} \frac{6-3x}{x+1}$ , direct substitution yields $\frac{6-3(2)}{2+1} = \frac{0}{3} = 0$ . * b) For the limit $\lim_{x \to 0} \frac{1-x+x^2-x^3}{3-x}$ , direct substitution yields $\frac{1}{3}$ . * c) For the limit $\lim_{x \to -4} \frac{16-x^2}{x^2+x-12}$ , the initial form is $\frac{0}{0}$ . Applying L'Hôpital's Rule: $\lim_{x \to -4} \frac{-2x}{2x+1} = \frac{-2(-4)}{2(-4)+1} = \frac{8}{-7} = -\frac{8}{7}$ . * d) For the limit $\lim_{x \to 4^-} (\frac{10}{x-4})^2$ , as $x \to 4^-$ , $\frac{10}{x-4} \to -\infty$ . Squaring this value leads to $\infty$ . * e) For the limit $\lim_{x \to 3^+} \frac{12-4x}{x-3}$ , the form is $\frac{0}{0}$ . Applying L'Hôpital's Rule: $\lim_{x \to 3^+} \frac{-4}{1} = -4$ . * f) For the limit $\lim_{x \to \infty} \frac{5x}{\ln(x)}$ , the form is $\frac{\infty}{\infty}$ . Applying L'Hôpital's Rule: $\lim_{x \to \infty} \frac{5}{1/x} = \lim_{x \to \infty} 5x = \infty$ . * g) For the limit $\lim_{x \to \infty} \frac{\ln(x)}{\ln(x^7)}$ , applying logarithmic properties yields $\lim_{x \to \infty} \frac{\ln(x)}{7\ln(x)} = \frac{1}{7}$ . * h) For the limit $\lim_{x \to \infty} (e^{-2x})$ , the value approaches $0$ . * i) For the limit $\lim_{x \to \infty} (\sin(x))$ , the result is DNE (Does Not Exist) because the function oscillates.

Derivative Calculations (Basic and Transcendental): * a) For $f(x) = \frac{5}{x} - 4x^6 + 3\sin^{-1}(x) + 3\sqrt[7]{x^2}$ , the derivative is $f'(x) = -5x^{-2} - 24x^5 + 3(\frac{1}{\sqrt{1-x^2}}) + \frac{6}{7}x^{-5/7}$ . * b) For $g(x) = \tan(e^{5x+2} + 7\ln(x))$ , the derivative is $g'(x) = [\sec^2(e^{5x+2} + 7\ln(x))] \times (5e^{5x+2} + \frac{7}{x})$ . * c) For $h(x) = (\sec(3x) + 8x)^{10}$ , the derivative is $h'(x) = 10(\sec(3x) + 8x)^9 \times (\sec(3x)\tan(3x) \times 3 + 8)$ . * d) For $F(t) = \frac{\ln(t^2 + 6t)}{\cos(t)}$ , the derivative using the quotient rule is $F'(t) = \frac{\cos(t)(\frac{2t+6}{t^2+6t}) - \ln(t^2+6t)(-\sin(t))}{\cos^2(t)}$ . * e) For $G(t) = e^{-2t}\tan^{-1}(4t)$ , the derivative using the product rule is $G'(t) = e^{-2t}(\frac{4}{1+(4t)^2}) + \tan^{-1}(4t) \times e^{-2t}(-2)$ .

Implicit Differentiation: * Given the equation $y^3 + x^3\sin(y) = 8$ , differentiate both sides with respect to $x$ : * $3y^2y' + x^3\cos(y)y' + 3x^2\sin(y) = 0$ * Factoring out $y'$ gives: $y'(3y^2 + x^3\cos(y)) = -3x^2\sin(y)$ * Solving for $y'$ : $y' = \frac{-3x^2\sin(y)}{3y^2 + x^3\cos(y)}$
Related Rates - Cube Volume: * Formula for volume of a cube: $V = s^3$ * Differentiate with respect to time $t$ : $\frac{dV}{dt} = 3s^2 \frac{ds}{dt}$ * Given: $\frac{dV}{dt} = 100\,in^3/sec$ and at the Moment of Interest (MOI), $s = 3\,in$ . * Substitution: $100 = 3(3)^2 \frac{ds}{dt} \rightarrow 100 = 27 \frac{ds}{dt}$ * Conclusion: $\frac{ds}{dt} = \frac{100}{27}\,in/sec$

Abstract Optimization: * Objective function: $f = a + b^2$ * Constraint: $a = 8 - 3b$ * Combined function: $f(b) = 8 - 3b + b^2$ * Derivative: $f'(b) = -3 + 2b$ . Setting to zero yields critical point at $b = \frac{3}{2}$ . * Solving for $a$ : $a = 8 - 3(\frac{3}{2}) = 8 - \frac{9}{2} = \frac{7}{2}$ .
Optimization of Fencing: * Fencing total length formula: $L = 2x + 4y$ . * Constraint based on area: $xy = 300 \rightarrow y = \frac{300}{x}$ . * Substitution for objective: $L(x) = 2x + 4(\frac{300}{x}) = 2x + \frac{1200}{x}$ . * Optimizing: $\frac{dL}{dx} = 2 - \frac{1200}{x^2} = 0 \rightarrow x^2 = 600 \rightarrow x = \sqrt{600}$ . * Minimal length value: $L(\sqrt{600}) = 2\sqrt{600} + \frac{1200}{\sqrt{600}} = 4\sqrt{600} = 40\sqrt{6}$ .

Linearization of a Radical Function: * Evaluate $f(x) = \sqrt{x^2+5}$ at $x=2$ . * $f(2) = \sqrt{4+5} = 3$ . * $f'(x) = \frac{x}{\sqrt{x^2+5}} \rightarrow f'(2) = \frac{2}{3}$ . * Linearization equation $L(x) = f(a) + f'(a)(x-a)$ : $L(x) = 3 + \frac{2}{3}(x-2)$ .
Linearization of Trigonometric Functions: * Linearize $f(x) = \tan(x)$ at $x = \frac{\pi}{4}$ . * $f(\frac{\pi}{4}) = 1$ and $f'(x) = \sec^2(x) \rightarrow f'(\frac{\pi}{4}) = 2$ . * Linearization equation: $L(x) = 1 + 2(x - \frac{\pi}{4})$ . * Approximating $\tan(\frac{6\pi}{25})$ : $L(\frac{6\pi}{25}) = 1 + 2(\frac{6\pi}{25} - \frac{\pi}{4}) = 1 + 2(\frac{24\pi - 25\pi}{100}) = 1 - \frac{\pi}{50}$ .

Given Derivative Information: * $f'(x) = \frac{5-x}{e^x}$ * $f''(x) = \frac{x-6}{e^x}$
Analytical Conclusions: * a) Increasing/Decreasing: $f'(x)$ changes sign at $x = 5$ . The function is decreasing on the interval $(5, \infty)$ . * b) Extrema: A local maximum exists at $x = 5$ ; there is no local minimum. * c) Concavity: $f''(x)$ changes sign at $x = 6$ . The function is concave down on $(-\infty, 6)$ . * d) Points of Inflection: By the second derivative test, the inflection point occurs at $x = 6$ .

Indefinite and Definite Integrals: * a) $\int (2x^4 + \frac{3}{x} - e^{-2x} + 2) \, dx = \frac{2}{5}x^5 + 3\ln|x| + \frac{1}{2}e^{-2x} + 2x + C$ * b) $\int (3\sec^2(x) - 2\cos(x) + \frac{5}{1+x^2}) \, dx = 3\tan(x) - 2\sin(x) + 5\tan^{-1}(x) + C$ * c) $\int_1^2 (\frac{4}{x} - 8x^3) \, dx = [4\ln|x| - 2x^4]_1^2 = (4\ln(2) - 32) - (0 - 2) = 4\ln(2) - 30$ * d) $\int_0^1 (2-e^{5x}) \, dx = [2x - \frac{1}{5}e^{5x}]_0^1 = (2 - \frac{1}{5}e^5) - (0 - \frac{1}{5}e^0) = 2 - \frac{e^5}{5} + \frac{1}{5} = \frac{11 - e^5}{5}$

Complex Forms: * a) $\int x^3(x^4+8)^{10} \, dx$ : Let $u = x^4+8, du = 4x^3dx \rightarrow \frac{1}{4} \int u^{10}du = \frac{1}{44}(x^4+8)^{11} + C$ . * b) $\int \frac{x^4}{10-3x^5} \, dx$ : Let $u = 10-3x^5, du = -15x^4dx \rightarrow -\frac{1}{15} \int \frac{1}{u}du = -\frac{1}{15}\ln|10-3x^5| + C$ . * c) $\int \sec^2(3x)e^{\tan(3x)} \, dx$ : Let $u = \tan(3x), du = 3\sec^2(3x)dx \rightarrow \frac{1}{3} \int e^u du = \frac{1}{3}e^{\tan(3x)} + C$ . * d) $\int_0^\pi (2-\sin(x))^4 \cos(x) \, dx$ : Let $u = 2-\sin(x), du = -\cos(x)dx$ . Lower limit $u(0)=2$ , upper limit $u(\pi)=2$ . The integral from $2$ to $2$ is $0$ .

Average Value: For $f(x) = x^2$ on the interval $[1, 4]$ : * $f_{avg} = \frac{1}{4-1} \int_1^4 x^2 \, dx = \frac{1}{3} [\frac{x^3}{3}]_1^4 = \frac{1}{9}(64-1) = \frac{63}{9} = 7$.
Absolute Extrema: For $f(x) = x^{2/3}$ on $[-8, 1]$ , the absolute maximum value is $f(-8) = 4$ .
The Fundamental Theorem of Calculus: $\frac{d}{dx} \int_7^x \sqrt{t^2+7} \, dt = \sqrt{x^2+7}$ .
Symmetry and Integrals: $\int_{-1}^1 x^{31} \, dx = 0$ because $x^{31}$ is an odd function over a symmetric interval.
Concept Verification (True/False): If $\lim_{x \to 4} f(x)$ exists, then $f$ is continuous at $x=4$ . (False).
Integral Algebra: Given $\int_1^5 g(x)dx = 4$ , $\int_1^8 g(x)dx = 1.5$ , and $\int_1^8 f(x)dx = 7$ , find: * i) $\int_5^8 g(x)dx = 1.5 - 4 = -2.5$ * ii) $\int_1^8 (f(x) - 3g(x))dx = 7 - 3(1.5) = 2.5$ * iii) $\int_8^1 f(x)dx = -7$

Total Accumulation Calculation: * Based on shaded regions with specified areas: Region 1 (below, area $2.0$ ), Region 2 (above, area $3.1$ ), Region 3 (below, area $6.2$ ), Region 4 (above, area $2.5$ ). * i) $\int_a^e f(x)dx = -2.0 + 3.1 - 6.2 + 2.5 = -2.6$ * ii) $\int_c^d f(x)dx = -6.2$ * iii) $\int_b^c f(x)dx = 3.1$ * iv) $\int_a^c f(x)dx = -2.0 + 3.1 = 1.1$