Calculus Study Summary: Differentiation and Integration

This summary covers key topics in differential and integral calculus, based on the provided questions. It includes various differentiation techniques (product rule, quotient rule, chain rule, implicit differentiation, logarithmic differentiation) and integration techniques (substitution, partial fractions, completing the square, integration by parts), as well as fundamental concepts of integrals.

I. Differentiation Techniques

1. Logarithmic Differentiation: Power of a Function

Question: Find \( \frac{dy}{dx} \) if \( y = x^{\tan x} \).

Method: Apply natural logarithm to both sides and differentiate implicitly.

Let \( y = x^{\tan x} \). Taking the natural logarithm of both sides: \( \ln y = \ln(x^{\tan x}) \) \( \ln y = \tan x \cdot \ln x \) Differentiate both sides with respect to \( x \): \( \frac{1}{y} \frac{dy}{dx} = (\sec^2 x \cdot \ln x) + (\tan x \cdot \frac{1}{x}) \) (using product rule on the right side) \( \frac{dy}{dx} = y \left( \sec^2 x \ln x + \frac{\tan x}{x} \right) \) Substitute \( y = x^{\tan x} \) back: \( \frac{dy}{dx} = x^{\tan x} \left( \frac{\tan x}{x} + \sec^2 x \ln x \right) \)

Correct Answer: \( \mathbf{x^{\tan x} \left( \frac{\tan x}{x} + \sec^2 x \ln x \right)} \)

2. Quotient Rule

Question: Find \( \frac{dy}{dx} \) if \( y = \frac{e^{\sin 2x}}{\cos 2x} \).

Method: Use the quotient rule \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).

Let \( u = e^{\sin 2x} \) and \( v = \cos 2x \). \( u' = e^{\sin 2x} \cdot (\cos 2x \cdot 2) = 2 \cos 2x \cdot e^{\sin 2x} \) \( v' = -\sin 2x \cdot 2 = -2 \sin 2x \) \( \frac{dy}{dx} = \frac{(2 \cos 2x \cdot e^{\sin 2x})(\cos 2x) - (e^{\sin 2x})(-2 \sin 2x)}{(\cos 2x)^2} \) Factor out \( e^{\sin 2x} \) from the numerator: \( \frac{dy}{dx} = \frac{e^{\sin 2x} (2 \cos^2 2x + 2 \sin 2x)}{\cos^2 2x} \) \( \frac{dy}{dx} = e^{\sin 2x} \left( \frac{2 \cos^2 2x}{\cos^2 2x} + \frac{2 \sin 2x}{\cos^2 2x} \right) \) \( \frac{dy}{dx} = e^{\sin 2x} (2 + 2 \tan 2x \sec 2x) \) \( \frac{dy}{dx} = 2 e^{\sin 2x} (1 + \sec 2x \tan 2x) \)

Correct Answer: \( \mathbf{2 e^{\sin 2x} (1 + \sec 2x \tan 2x)} \)

3. Product Rule (with Natural Logarithm)

Question: If \( y = e^{2x} \log(x+6) \), find \( \frac{dy}{dx} \) (assuming \( \log \) is natural logarithm, \( \ln \)).

Method: Use the product rule \( (uv)' = u'v + uv' \).

Let \( u = e^{2x} \) and \( v = \ln(x+6) \). \( u' = 2e^{2x} \) \( v' = \frac{1}{x+6} \) \( \frac{dy}{dx} = (2e^{2x})\ln(x+6) + e^{2x}\left(\frac{1}{x+6}\right) \) Factor out \( e^{2x} \): \( \frac{dy}{dx} = e^{2x} \left( 2\ln(x+6) + \frac{1}{x+6} \right) \)

Correct Answer: \( \mathbf{e^{2x} \left( \frac{1}{x+6} + 2 \ln(x+6) \right)} \)

4. Implicit Differentiation

Question: Differentiate \( x^2 \sin y - e^{2y} \cos x = 3x^2 + 2x - 5 \) with respect to \( x \).

Method: Differentiate each term with respect to \( x \), remembering to use the chain rule for terms involving \( y \) (multiplying by \( \frac{dy}{dx} \)).

Differentiating \( x^2 \sin y \): \( 2x \sin y + x^2 \cos y \frac{dy}{dx} \) Differentiating \( -e^{2y} \cos x \): \( -\left( (2e^{2y} \frac{dy}{dx})\cos x + e^{2y}(-\sin x) \right) = -2e^{2y} \cos x \frac{dy}{dx} + e^{2y} \sin x \) Differentiating \( 3x^2 + 2x - 5 \): \( 6x + 2 \) Combine the terms: \( 2x \sin y + x^2 \cos y \frac{dy}{dx} - 2e^{2y} \cos x \frac{dy}{dx} + e^{2y} \sin x = 6x + 2 \) Group terms with \( \frac{dy}{dx} \): \( \frac{dy}{dx} (x^2 \cos y - 2e^{2y} \cos x) = 6x + 2 - 2x \sin y - e^{2y} \sin x \) Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{6x + 2 - 2x \sin y - e^{2y} \sin x}{x^2 \cos y - 2e^{2y} \cos x} \)

Correct Answer: \( \mathbf{\frac{6x + 2 - 2x \sin y - e^{2y} \sin x}{x^2 \cos y - 2e^{2y} \cos x}} \)

5. Logarithm Properties and Chain Rule

Question: Obtain \( \frac{d}{dx} \left\{ \ln\left(\frac{x^2}{x^2 + 1}\right) \right\} \).

Method: Use logarithm properties \( \ln\left(\frac{A}{B}\right) = \ln A - \ln B \) before differentiating.

Let \( y = \ln\left(\frac{x^2}{x^2 + 1}\right) = \ln(x^2) - \ln(x^2 + 1) = 2\ln x - \ln(x^2 + 1) \) Differentiate with respect to \( x \): \( \frac{dy}{dx} = 2 \cdot \frac{1}{x} - \frac{1}{x^2 + 1} \cdot (2x) \) \( \frac{dy}{dx} = \frac{2}{x} - \frac{2x}{x^2 + 1} \) Combine into a single fraction: \( \frac{dy}{dx} = \frac{2(x^2 + 1) - 2x(x)}{x(x^2 + 1)} = \frac{2x^2 + 2 - 2x^2}{x(x^2 + 1)} = \frac{2}{x(x^2 + 1)} \)

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

6. Differentiation of Basic Logarithmic Function

Question: Find \( \frac{dy}{dx} \) if \( y = \log_e \cos x \).

Method: Use the chain rule for \( \ln(f(x)) \).

Given \( y = \ln(\cos x) \) \( \frac{dy}{dx} = \frac{1}{\cos x} \cdot (-\sin x) = -\frac{\sin x}{\cos x} = -\tan x \)

Correct Answer: \( \mathbf{-\tan x} \)

7. Differentiation of Mixed Log and Exponential Power Function

Question: Differentiate \( y = \log(x^3 - x) + 7^{\frac{3}{2}x} \) (assuming \( \log \) is natural logarithm for the first term).

Method: Differentiate each term separately using chain rule and power rules.

For the first term, \( \frac{d}{dx}(\ln(x^3 - x)) = \frac{1}{x^3 - x} \cdot (3x^2 - 1) = \frac{3x^2 - 1}{x(x^2 - 1)} \). For the second term, \( \frac{d}{dx}(a^{kx}) = k \cdot a^{kx} \cdot \ln a \). Here \( a=7, k=\frac{3}{2} \). So, \( \frac{d}{dx}(7^{\frac{3}{2}x}) = \frac{3}{2} \cdot 7^{\frac{3}{2}x} \cdot \ln 7 \). Combining them: \( \frac{dy}{dx} = \frac{3x^2 - 1}{x(x^2 - 1)} + \frac{3}{2} \cdot 7^{\frac{3}{2}x} \cdot \ln 7 \).

Note on Options: The options presented in the document show a slightly different constant factor for the second term, e.g., \( 7^{\frac{3}{2}x} (\frac{2}{3} \log 7) \), which is mathematically inconsistent with the standard derivative. However, matching the form, the option with correct first term and positive exponential part is generally preferred.

Closest Answer (based on structure and ignoring potential constant typo): \( \mathbf{\frac{3x^2 - 1}{x(x^2 - 1)} + 7^{\frac{3}{2}x} (\text{constant factor})} \)

8. Implicit Differentiation (Polynomial Form)

Question: Find \( \frac{dy}{dx} \) if \( 3x^2 y + xy^2 = 5 \).

Method: Differentiate each term implicitly with respect to \( x \).

Differentiating \( 3x^2 y \): \( 6xy + 3x^2 \frac{dy}{dx} \) Differentiating \( xy^2 \): \( y^2 + x(2y \frac{dy}{dx}) = y^2 + 2xy \frac{dy}{dx} \) Differentiating \( 5 \): \( 0 \) Combine: \( 6xy + 3x^2 \frac{dy}{dx} + y^2 + 2xy \frac{dy}{dx} = 0 \) Group terms with \( \frac{dy}{dx} \): \( \frac{dy}{dx} (3x^2 + 2xy) = -6xy - y^2 \) Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{-6xy - y^2}{3x^2 + 2xy} = -\frac{y^2 + 6xy}{3x^2 + 2xy} \)

Correct Answer: \( \mathbf{-\frac{y^2 + 6xy}{3x^2 + 2xy}} \)

II. Integration Techniques

1. Integration by Substitution

Question: Evaluate \( \int \frac{\cos(\ln x)}{x} dx \).

Method: Use \( u \)-substitution.

Let \( u = \ln x \). Then \( du = \frac{1}{x} dx \). The integral becomes \( \int \cos u \, du \) \( \int \cos u \, du = \sin u + C \) Substitute back \( u = \ln x \): \( \sin(\ln x) + C \)

Correct Answer: \( \mathbf{\sin(\ln x) + C} \)

2. Integration by Partial Fractions

Question: Simplify \( \int \frac{4x - 5}{x^2 - x - 2} dx \).

Method: Factor the denominator and use partial fraction decomposition.

Factor the denominator: \( x^2 - x - 2 = (x - 2)(x + 1) \). Set up partial fractions: \( \frac{4x - 5}{(x - 2)(x + 1)} = \frac{A}{x - 2} + \frac{B}{x + 1} \) Multiply by \( (x - 2)(x + 1) \): \( 4x - 5 = A(x + 1) + B(x - 2) \) Set \( x = 2 \): \( 4(2) - 5 = A(2 + 1) \implies 3 = 3A \implies A = 1 \) Set \( x = -1 \): \( 4(-1) - 5 = B(-1 - 2) \implies -9 = -3B \implies B = 3 \) So the integral becomes \( \int \left( \frac{1}{x - 2} + \frac{3}{x + 1} \right) dx \) Integrate each term: \( \ln|x - 2| + 3\ln|x + 1| + C \)

Correct Answer: \( \mathbf{\ln|x - 2| + 3\ln|x + 1| + C} \)

3. Integration by Substitution (Hyperbolic Form)

Question: Evaluate \( \int \frac{e^{2x} + e^{-2x}}{e^{2x} - e^{-2x}} dx \).

Method: Use \( u \)-substitution.

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 \( u = e^{2x} - e^{-2x} \): \( \frac{1}{2} \ln|e^{2x} - e^{-2x}| + C \)

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

4. Integration by Completing the Square

Question: Evaluate \( \int \frac{dx}{x^2 + 4x + 7} \).

Method: Complete the square in the denominator to use the arctangent integral form \( \int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C \).

Complete the square for \( x^2 + 4x + 7 \): \( x^2 + 4x + 4 + 3 = (x + 2)^2 + 3 \) Let \( u = x + 2 \), so \( du = dx \). Let \( a^2 = 3 \), so \( a = \sqrt{3} \). The integral becomes \( \int \frac{du}{u^2 + (\sqrt{3})^2} \) \( \frac{1}{\sqrt{3}} \arctan\left(\frac{u}{\sqrt{3}}\right) + C \) Substitute back \( u = x + 2 \): \( \frac{1}{\sqrt{3}} \arctan\left(\frac{x + 2}{\sqrt{3}}\right) + C \)

Correct Answer: \( \mathbf{\frac{1}{\sqrt{3}} \arctan\left(\frac{x + 2}{\sqrt{3}}\right) + C} \)

5. Integration by Parts (for \( \ln x \))

Question: Simplify \( \int \ln x \, dx \).

Method: Use integration by parts \( \int u \, dv = uv - \int v \, du \).

Let \( u = \ln x \) and \( dv = dx \). Then \( du = \frac{1}{x} dx \) and \( v = x \). \( \int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} dx \) \( = x \ln x - \int 1 \, dx \) \( = x \ln x - x + C \)

Correct Answer: \( \mathbf{x \ln x - x + C} \)

6. Integration by Parts (Cyclic Integral)

Question: Simplify \( \int e^x \sin x \, dx \).

Method: Use integration by parts twice, leading to a cyclic equation.

Let \( I = \int e^x \sin x \, dx \). First application of integration by parts: Let \( u = \sin x \), \( dv = e^x dx \). Then \( du = \cos x dx \), \( v = e^x \). \( I = e^x \sin x - \int e^x \cos x \, dx \) Second application of integration by parts (on \( \int e^x \cos x \, dx \)): Let \( u = \cos x \), \( dv = e^x dx \). Then \( du = -\sin x dx \), \( v = e^x \). \( \int e^x \cos x \, dx = e^x \cos x - \int e^x (-\sin x) dx = e^x \cos x + \int e^x \sin x dx \) Substitute this back into the equation for \( I \): \( I = e^x \sin x - (e^x \cos x + I) \) \( I = e^x \sin x - e^x \cos x - I \) \( 2I = e^x (\sin x - \cos x) \) \( I = \frac{e^x}{2} (\sin x - \cos x) + C \)

Correct Answer: \( \mathbf{\frac{e^x}{2} (\sin x - \cos x) + C} \)

III. Fundamental Concepts

1. Properties of Definite Integrals

Question: Which of the following equations is not correct?

• \( \int_a^a f(x)dx = 0 \) (Correct: The integral over an interval of zero length is zero.)
• \( \int_a^b f(x)dx = \int_b^a f(x)dx \) (Incorrect: The limits of integration are reversed, so the sign should change. The correct property is \( \int_a^b f(x)dx = -\int_b^a f(x)dx \).)
• \( \left|\int_a^b f(x)dx\right| \le \int_a^b |f(x)|dx \) if \( a < b \) (Correct: Triangle Inequality for Integrals.)
• \( \int_a^b f(x)dx \le \int_a^b g(x)dx \) if \( f(x) \le g(x) \) in \( [a,b] \) (Correct: Comparison Property of Integrals.)

Correct Answer: The incorrect statement is \( \mathbf{\int_a^b f(x)dx = \int_b^a f(x)dx} \).

2. Definition of Indefinite Integral

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

Explanation: The expression \( \int f(x)dx \) represents the family of all antiderivatives of \( f(x) \), differing by a constant. This is known as the indefinite integral.

Correct Answer: Indefinite integral

This study summary provides a concise overview of the calculus problems presented in the document, outlining the methods and correct solutions for each type of question.