MTH 102 Test Compilation 9

Comprehensive Calculus Study Summary

This document covers a range of calculus topics including differentiation (explicit, implicit, logarithmic), integration (substitution, partial fractions, trigonometric), and properties/definitions related to integrals.

I. Differentiation

This section summarizes various differentiation problems encountered.

1. Derivatives of Logarithmic Functions

• Problem: Obtain $\frac{d}{dx} \left\{ \ln\left(\frac{x^2}{x^2 + 1}\right) \right\}$

  Solution: Using logarithm properties, $\ln\left(\frac{x^2}{x^2 + 1}\right) = \ln(x^2) - \ln(x^2 + 1) = 2\ln(x) - \ln(x^2 + 1)$.

  Differentiating term by term: $\frac{d}{dx} (2\ln x) = \frac{2}{x}$ and $\frac{d}{dx} (\ln(x^2 + 1)) = \frac{2x}{x^2 + 1}$.

  Combining and simplifying: $\frac{2}{x} - \frac{2x}{x^2 + 1} = \frac{2(x^2 + 1) - 2x^2}{x(x^2 + 1)} = \frac{2x^2 + 2 - 2x^2}{x(x^2 + 1)} = \frac{2}{x(x^2 + 1)}$.

  Correct Answer: $\frac{2}{x(x^2 + 1)}$

• Problem: Differentiate $y = \log(x^3 - x) + 7^{3x}$

  Note on potential typo: Based on the options, it is highly probable that $\log$ denotes the natural logarithm ($\ln$), and the exponential term in the original function should have been $7^{\frac{2}{3}x}$ instead of $7^{3x}$, with the options having a typo in the exponent displaying $7^{3x}$ while using $\frac{2}{3}\log 7$ for the derivative coefficient.

  Solution (assuming $y = \ln(x^3 - x) + 7^{\frac{2}{3}x}$ and $\log 7 = \ln 7$):

  • For the first term: $\frac{d}{dx} (\ln(x^3 - x)) = \frac{3x^2 - 1}{x^3 - x} = \frac{3x^2 - 1}{x(x^2 - 1)}$.
  • For the second term: $\frac{d}{dx} (7^{\frac{2}{3}x}) = 7^{\frac{2}{3}x} \cdot \ln(7) \cdot \frac{2}{3}$.

  Combined Derivative: $\frac{3x^2 - 1}{x(x^2 - 1)} + \frac{2}{3} \cdot 7^{\frac{2}{3}x} \cdot \ln(7)$.

  Correct Answer (closest match due to assumed typo): $\frac{3x^2 - 1}{x(x^2 - 1)} + 7^{3x} \left(\frac{2}{3} \log 7\right)$

• Problem: Find $\frac{dy}{dx}$, if $y = \log_e (\cos x)$

  Solution: Using the chain rule, $\frac{d}{dx} (\ln(u)) = \frac{u'}{u}$. Here, $u = \cos x$, so $u' = -\sin x$.

  $\frac{dy}{dx} = \frac{-\sin x}{\cos x} = -\tan x$.

  Correct Answer: $-\tan x$

2. Implicit Differentiation

• Problem: Differentiate $x^2 \sin y - e^{2y} \cos x = 3x^2 + 2x - 5$ with respect to x.

  Solution: Differentiate each term using product and chain rules:

  • $\frac{d}{dx} (x^2 \sin y) = 2x \sin y + x^2 \cos y \frac{dy}{dx}$
  • $\frac{d}{dx} (-e^{2y} \cos x) = -( (e^{2y} \cdot 2 \frac{dy}{dx}) \cos x + e^{2y} (-\sin x) ) = -2e^{2y} \cos x \frac{dy}{dx} + e^{2y} \sin x$
  • $\frac{d}{dx} (3x^2 + 2x - 5) = 6x + 2$

  Combining terms and isolating $\frac{dy}{dx}$:

  $2x \sin y + x^2 \cos y \frac{dy}{dx} - 2e^{2y} \cos x \frac{dy}{dx} + e^{2y} \sin x = 6x + 2$

  $\frac{dy}{dx} (x^2 \cos y - 2e^{2y} \cos x) = 6x + 2 - 2x \sin y - e^{2y} \sin x$

  $\frac{dy}{dx} = \frac{6x - 2x \sin y - e^{2y} \sin x + 2}{x^2 \cos y - 2e^{2y} \cos x}$

  Correct Answer: $\frac{dy}{dx} = \frac{6x - 2x \sin y - \sin x e^{2y} + 2}{x^2 \cos y - 2 \cos x e^{2y}}$

• Problem: Find $\frac{dy}{dx}$ if $3x^2 y + xy^2 = 5$

  Solution: Differentiate implicitly with respect to x:

  • $\frac{d}{dx} (3x^2 y) = 6xy + 3x^2 \frac{dy}{dx}$
  • $\frac{d}{dx} (xy^2) = y^2 + x (2y \frac{dy}{dx}) = y^2 + 2xy \frac{dy}{dx}$
  • $\frac{d}{dx} (5) = 0$

  Combining and solving for $\frac{dy}{dx}$:

  $6xy + 3x^2 \frac{dy}{dx} + y^2 + 2xy \frac{dy}{dx} = 0$

  $\frac{dy}{dx} (3x^2 + 2xy) = -6xy - y^2$

  $\frac{dy}{dx} = \frac{-y^2 - 6xy}{3x^2 + 2xy} = -\frac{y^2 + 6xy}{3x^2 + 2xy}$

  Correct Answer: $\frac{dy}{dx} = -\frac{y^2 + 6xy}{3x^2 + 2xy}$

3. Derivatives of Exponential and Power Functions

• Problem: Find $\frac{dy}{dx}$ if $y = e^{2x} \log(x+6)$

  Solution: Using the product rule $\frac{d}{dx}(uv) = u'v + uv'$, where $u = e^{2x}$ and $v = \log(x+6)$ (assuming natural logarithm $\ln$).

  • $u' = \frac{d}{dx}(e^{2x}) = 2e^{2x}$
  • $v' = \frac{d}{dx}(\ln(x+6)) = \frac{1}{x+6}$

  $\frac{dy}{dx} = (2e^{2x}) \ln(x+6) + e^{2x} \left(\frac{1}{x+6}\right) = e^{2x} \left[2\ln(x+6) + \frac{1}{x+6}\right]$.

  Correct Answer: $\frac{dy}{dx} = e^{2x} \left[\frac{1}{x+6} + 2\log(x+6)\right]$

• Problem: Find $\frac{dy}{dx}$ if $y = x^{\tan x}$

  Solution: Use logarithmic differentiation.

  Take natural logarithm of both sides: $\ln y = \ln (x^{\tan x}) = (\tan x) \ln x$.

  Differentiate implicitly with respect to x:

  $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (\tan x \cdot \ln x)$

  Using product rule: $\frac{d}{dx} (\tan x \cdot \ln x) = (\sec^2 x) \ln x + (\tan x) \left(\frac{1}{x}\right)$.

  $\frac{1}{y} \frac{dy}{dx} = (\sec^2 x) \ln x + \frac{\tan x}{x}$

  $\frac{dy}{dx} = y \left[ (\sec^2 x) \ln x + \frac{\tan x}{x} \right] = x^{\tan x} \left[ \frac{\tan x}{x} + (\sec^2 x) \ln x \right]$.

  Correct Answer: $\frac{dy}{dx} = x^{\tan x} \left[ \frac{\tan x}{x} + \sec^2 x \ln x \right]$

• Problem: Find $\frac{dy}{dx}$ if $y = \frac{e^{\sin 2x}}{\cos 2x}$

  Solution: Use the quotient rule or product rule with negative exponent. Let's use the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$.

  • $u = e^{\sin 2x} \implies u' = e^{\sin 2x} \cdot (\cos 2x \cdot 2) = 2\cos 2x \cdot e^{\sin 2x}$
  • $v = \cos 2x \implies v' = -\sin 2x \cdot 2 = -2\sin 2x$

  $\frac{dy}{dx} = \frac{(2\cos 2x \cdot e^{\sin 2x})(\cos 2x) - (e^{\sin 2x})(-2\sin 2x)}{(\cos 2x)^2}$

  $\frac{dy}{dx} = \frac{2\cos^2 2x \cdot e^{\sin 2x} + 2\sin 2x \cdot e^{\sin 2x}}{\cos^2 2x}$

  $\frac{dy}{dx} = 2e^{\sin 2x} \frac{\cos^2 2x + \sin 2x}{\cos^2 2x} = 2e^{\sin 2x} \left(1 + \frac{\sin 2x}{\cos^2 2x}\right)$

  $\frac{dy}{dx} = 2e^{\sin 2x} \left(1 + \frac{\sin 2x}{\cos 2x} \cdot \frac{1}{\cos 2x}\right) = 2e^{\sin 2x} (1 + \tan 2x \sec 2x)$.

  Correct Answer: $2e^{\sin 2x} [1 + \sec 2x \tan 2x]$

II. Integration

This section outlines various integration problems.

1. Integration by Substitution

• Problem: Simplify $\int 3x^2 e^{x^3} dx$

  Solution: Let $u = x^3$. Then $du = 3x^2 dx$. The integral becomes:

  $\int e^u du = e^u + C$.

  Substitute back: $e^{x^3} + C$.

  Correct Answer: $e^{x^3} + C$

• Problem: Evaluate $\int \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}} dx$

  Solution: Let $u = e^{2x} - e^{-2x}$. Then $du = (2e^{2x} - (-2e^{-2x})) dx = (2e^{2x} + 2e^{-2x}) dx = 2(e^{2x} + e^{-2x}) dx$.

  So, $(e^{2x} + e^{-2x}) dx = \frac{1}{2} du$. The integral becomes:

  $\int \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| + C$.

  Substitute back: $\frac{1}{2} \ln|e^{2x} - e^{-2x}| + C$.

  Correct Answer: $\frac{1}{2} \ln|e^{2x} - e^{-2x}| + C$

2. Integration with Trigonometric Functions

• Problem: Evaluate $\int \cos^3 x dx$

  Solution: Rewrite $\cos^3 x$ as $\cos^2 x \cdot \cos x$. Use the identity $\cos^2 x = 1 - \sin^2 x$.

  $\int (1 - \sin^2 x) \cos x dx$.

  Let $u = \sin x$. Then $du = \cos x dx$. The integral becomes:

  $\int (1 - u^2) du = u - \frac{u^3}{3} + C$.

  Substitute back: $\sin x - \frac{\sin^3 x}{3} + C$.

  Correct Answer: $\sin x - \frac{1}{3}\sin^3 x + C$

3. Integration with Partial Fractions

• Problem: Simplify $\int \frac{4x - 5}{x^2 - x - 2} dx$

  Solution: First, factor the denominator: $x^2 - x - 2 = (x - 2)(x + 1)$.

  Decompose the integrand into partial fractions: $\frac{4x - 5}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1}$.

  $4x - 5 = A(x + 1) + B(x - 2)$.

  • Set $x = 2$: $4(2) - 5 = A(2 + 1) \implies 3 = 3A \implies A = 1$.
  • Set $x = -1$: $4(-1) - 5 = B(-1 - 2) \implies -9 = -3B \implies B = 3$.

  The integral becomes: $\int \left(\frac{1}{x - 2} + \frac{3}{x + 1}\right) dx$.

  Integrate each term: $\int \frac{1}{x - 2} dx + \int \frac{3}{x + 1} dx = \ln|x - 2| + 3\ln|x + 1| + C$.

  Correct Answer: $3\ln|x+1| + \ln|x-2| + C$

4. Integration by Completing the Square (Inverse Trigonometric Form)

• Problem: Evaluate $\int \frac{dx}{x^2 + 4x + 7}$

  Solution: Complete the square in the denominator: $x^2 + 4x + 7 = (x^2 + 4x + 4) + 3 = (x + 2)^2 + (\sqrt{3})^2$.

  The integral is of the form $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$.

  Here, $u = x + 2$ (so $du = dx$) and $a = \sqrt{3}$.

  $\int \frac{dx}{(x + 2)^2 + (\sqrt{3})^2} = \frac{1}{\sqrt{3}} \arctan\left(\frac{x + 2}{\sqrt{3}}\right) + C$.

  Correct Answer: $\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{x + 2}{\sqrt{3}}\right) + C$

III. Properties and Definitions of Integrals

1. Properties of Definite Integrals

• Problem: Which of the following equations is not correct?
  1. $\int_a^a f(x)dx = 0$ (Correct)
  2. $\int_a^b f(x)dx = \int_b^a f(x)dx$ (Incorrect)
  3. $|\int_a^b f(x)dx| \leq \int_a^b |f(x)|dx$ if $a < b$ (Correct, Triangle Inequality for Integrals)
  4. $\int_a^b f(x)dx \leq \int_a^b g(x)dx$ if $f(x) \leq g(x)$ in $[a,b]$ (Correct, Comparison Property of Integrals)

  Explanation: The fundamental property states that $\int_a^b f(x)dx = -\int_b^a f(x)dx$. Therefore, option 'b' which omits the negative sign is incorrect.

  Correct Answer: $\int_a^b f(x)dx = \int_b^a f(x)dx$

2. Definitions in Integration

• Problem: If $y = \int f(x)dx + C$, then $f(x)$ is called the -----.

  Explanation: In indefinite integration, the function $f(x)$ that is being integrated is called the integrand.

  Correct Answer: integrand