Calculus

Meaning

Imagine walking along a path without ever lifting your feet from the ground. If you can trace a graph from left to right on your paper without once picking up your pencil, that graph is continuous. In plain English, continuity means there are no sudden jumps, breaks, or holes in a function.

Definitions

1. According to mathematician Augustin-Louis Cauchy, a function is continuous if an infinitely small change in the variable produces an infinitely small change in the value of the function itself.
2. In modern textbook terms, a function \(f(x)\) is continuous at a specific point \(x = c\) if the limit of the function as \(x\) approaches \(c\) is equal to the value of the function at \(c\). Mathematically, this is written as: $$\lim_{x \to c} f(x) = f(c)$$

Explanation

• To prove continuity at a point \(x = c\), three things must happen: The function \(f(c)\) must be defined; the limit of \(f(x)\) as \(x\) approaches \(c\) must exist (meaning Left-Hand Limit equals Right-Hand Limit: \(\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)\)); and the limit value must equal the function value at that point.
• Polynomial functions, trigonometric functions (within their domains), exponential functions, and logarithmic functions are continuous everywhere in their respective domains.
• Greatest integer functions and step functions have breaks and are discontinuous at integer values.

Examples

• \(f(x) = x^2 + 3x\) is continuous everywhere because you can plug in any real number without getting an undefined value.
• \(f(x) = \frac{1}{x-2}\) is not continuous at \(x = 2\) because the denominator becomes zero, creating a vertical asymptote (a break in the graph).

Real-Life Use & Misconceptions

Real-Life Use: Think about driving a car on a smooth highway. Your distance from your starting point increases continuously with time. You cannot magically teleport from mile marker ten to mile marker twenty without covering the distance in between. Temperature changes throughout the day also follow a continuous curve.

Common Misconception: If a graph looks smooth, it is always continuous.
Correction: Not necessarily. A graph can have a missing point (a hole) that you might not notice at first glance, making it discontinuous at that exact point.

Meaning

Differentiability tells us whether a function has a sharp corner or not. If continuity is about not lifting your pencil, differentiability is about not having any sharp spikes or sudden turns. A differentiable curve is completely smooth.

Definitions

1. According to calculus pioneer Gottfried Wilhelm Leibniz, differentiability relates to the existence of a unique tangent line at every point along a curve.
2. Formally, a function \(f(x)\) is differentiable at \(x = c\) if the derivative limit exists finitely: $$f'(c) = \lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$$

Explanation

• For a function to be differentiable at a point, its Left-Hand Derivative (LHD) must equal its Right-Hand Derivative (RHD) at that point.
• Golden Rule: Differentiability implies continuity, but continuity does NOT imply differentiability. (Think of the absolute value function \(f(x) = |x|\) at \(x = 0\). It is continuous there, but it has a sharp "V" shape, so it is not differentiable).
• Sharp corners, cusps, vertical tangents, and points of discontinuity destroy differentiability, whereas smooth, flowing curves are always differentiable.

Examples

• \(f(x) = x^2\) is differentiable everywhere because its derivative \(f'(x) = 2x\) exists for all real numbers.
• \(f(x) = |x|\) is continuous at \(x = 0\), but its derivative does not exist at \(x = 0\) due to the sharp vertex.

Real-Life Use & Misconceptions

Real-Life Use: Roller coaster design! Engineers use differentiability to ensure that the tracks transition smoothly without sharp angles. If a track is not differentiable, the sudden change in acceleration would cause whiplash or structural failure.

Common Misconception: If a function is continuous, it must be differentiable everywhere.
Correction: Remember the absolute value function. It is continuous everywhere, but fails to be differentiable at the origin because of the sharp corner.

Meaning

These are specialized techniques and formulas to find rates of change for complex mathematical structures.

Definitions

1. The Chain Rule (for composite functions): If \(y = f(g(x))\), then the derivative is \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\).
2. Implicit Differentiation: A method used when a function is not explicitly solved for \(y\) in terms of \(x\) (such as \(x^2 + y^2 = 25\)), allowing us to differentiate both sides with respect to \(x\) while treating \(y\) as a function of \(x\).

Explanation

• Composite Functions: Peel the function like an onion. Differentiate the outer layer first while leaving the inner layer untouched, then multiply by the derivative of the inner layer.
• Inverse Trig Functions: Memorize standard formulas like \(\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}\).
• Exponential & Logarithmic: \(\frac{d}{dx}(e^x) = e^x\), and \(\frac{d}{dx}(\ln x) = \frac{1}{x}\).
• Always look out for the chain rule when dealing with nested functions (like \(\sin(x^2)\)). When differentiating implicit equations containing \(y\), always attach \(\frac{dy}{dx}\) after differentiating any term containing \(y\).

Examples

• For \(y = (3x + 5)^4\), using the chain rule: \(\frac{dy}{dx} = 4(3x + 5)^3 \cdot 3 = 12(3x + 5)^3\).
• For \(y = e^{2x}\), the derivative is \(\frac{dy}{dx} = 2e^{2x}\).

Real-Life Use

In economics, composite functions are used to calculate compound interest where the interest rate itself depends on time and inflation simultaneously. Exponential derivatives model radioactive decay and population growth.

Meaning

• Logarithmic Differentiation: Taking the natural logarithm on both sides of an equation to simplify complex products, quotients, or expressions raised to the power of another variable (functions of the form \(f(x)^{g(x)}\)).
• Parametric Forms: When both \(x\) and \(y\) are expressed in terms of a third variable (called a parameter, usually \(t\) or \(\theta\)), such as \(x = \cos(t)\) and \(y = \sin(t)\).

Explanation

• If you see a function raised to another function (like \(x^x\)), normal power rules and exponential rules fail. Taking \(\ln\) brings the exponent down using logarithm laws (\(\ln(a^b) = b \ln a\)).
• For parametric equations, find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) separately, then divide them to get the derivative: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

Examples

• To find the derivative of \(y = x^x\): take \(\ln\) on both sides to get \(\ln y = x \ln x\), then differentiate implicitly.
• If \(x = 2t\) and \(y = 3t^2\), then \(\frac{dx}{dt} = 2\) and \(\frac{dy}{dt} = 6t\). Thus, \(\frac{dy}{dx} = \frac{6t}{2} = 3t\).

Real-Life Use

Parametric equations are heavily used in computer animation and physics to track projectile motion where both horizontal and vertical positions depend independently on time.

Meaning

The derivative of a derivative! If the first derivative tells you the rate of change (velocity), the second derivative tells you the rate of change of that rate of change (acceleration).

Explanation

• Denoted as \(\frac{d^2y}{dx^2}\) or \(f''(x)\).
• You simply differentiate the original function once to get \(f'(x)\), and then differentiate that result again with respect to \(x\).

Examples

• If \(f(x) = 4x^3 - 5x^2 + 7\), then \(f'(x) = 12x^2 - 10x\), and the second derivative is \(f''(x) = 24x - 10\).

Real-Life Use

In physics, if position is \(s(t)\), velocity is \(s'(t)\), and acceleration is \(s''(t)\). If you are in a car, the second derivative represents how abruptly you are slamming on the gas pedal or brakes (physicists call the rate of change of acceleration "jerk").

Meaning

These are foundational theorems in calculus that guarantee the existence of a specific point on a curve where the tangent line behaves in a particular way.

Definitions

1. Rolle's Theorem: If a function \(f(x)\) is continuous on \([a, b]\), differentiable on \((a, b)\), and \(f(a) = f(b)\), then there exists at least one number \(c\) in \((a, b)\) such that \(f'(c) = 0\) (meaning the tangent is horizontal).
2. Lagrange's Mean Value Theorem: If \(f(x)\) is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists at least one number \(c\) in \((a, b)\) such that: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$

Explanation

• Imagine throwing a ball straight up in the air and catching it at the exact same height. At the very peak of its flight, its vertical velocity is zero for an instant. That is Rolle's Theorem in action.
• MVT states that the instantaneous rate of change at some point inside an interval equals the average rate of change over the entire interval.

Examples

• \(f(x) = x^2\) on \([-2, 2]\). Here \(f(-2) = 4\) and \(f(2) = 4\). By Rolle's theorem, there must be a point where \(f'(c) = 0\). Since \(f'(x) = 2x\), setting \(2c = 0\) gives \(c = 0\), which is inside the interval.

Real-Life Use

Highway patrol speed enforcement! If you drive a distance of \(100\) miles on a toll road in one hour, your average speed is \(100\) miles per hour. By the Mean Value Theorem, your speedometer must have hit at least \(100\) miles per hour at some exact instant during that trip, even if you stopped at a rest stop.

Meaning

This is where calculus meets the real world. We use derivatives to optimize business profits, find maximum speeds, calculate slopes of curves, and approximate complex numbers.

Explanation

• Rate of Change: \(\frac{dy}{dx}\) tells us how fast quantity \(y\) changes with respect to \(x\).
• Increasing/Decreasing: If \(f'(x) > 0\), the function is increasing (going up). If \(f'(x) < 0\), the function is decreasing (going down).
• Tangents and Normals: The slope of the tangent line to a curve at a point is \(m = f'(x_0)\). The normal line is perpendicular to the tangent, so its slope is \(-\frac{1}{f'(x_0)}\).
• Maxima and Minima: To find peak profits or minimum costs, find where \(f'(x) = 0\) (critical points). Use the Second Derivative Test (\(f''(x) < 0\) for maximum, \(f''(x) > 0\) for minimum).

Examples

• A company's profit function is \(P(x) = 100x - x^2\). To find the production level \(x\) that maximizes profit, take the derivative: \(P'(x) = 100 - 2x\). Set it to zero: \(100 - 2x = 0 \implies x = 50\) units.

Real-Life Use

Every tech company, manufacturing plant, and financial institution uses optimization daily. Amazon uses derivatives to find the minimum shipping cost routes; pharmaceutical companies use them to find the optimal dosage of medicine that maximizes healing while minimizing toxic side effects.

Meaning

If derivatives are about breaking things down into rates of change, integration is about adding everything back up. Integration is the reverse process of differentiation (often called anti-differentiation).

Definitions

1. The Fundamental Theorem of Calculus (FTC): Links the concept of the derivative with the concept of the integral. If \(f\) is continuous on \([a, b]\) and \(F\) is any anti-derivative of \(f\), then: $$\int_{a}^{b} f(x) dx = F(b) - F(a)$$

Explanation

• If differentiating \(x^3\) gives \(3x^2\), then integrating \(3x^2\) gives back \(x^3\) (plus a constant of integration \(+ C\) because the derivative of any constant is zero).
• Geometric Interpretation: Definite integrals calculate the exact net area bounded by a curve and the horizontal axis between two limits \(a\) and \(b\).

Examples

• \(\int 4x^3 dx = x^4 + C\).
• \(\int_{0}^{2} 2x dx = [x^2]_0^2 = (2^2) - (0^2) = 4\).

Real-Life Use

If you know the velocity of a rocket at every single second, integration allows you to calculate the exact total distance the rocket traveled. In civil engineering, integrating load curves tells architects the total weight a bridge beam must support.

Meaning

Techniques used to solve integrals that cannot be integrated using basic standard formulas.

Explanation

• Substitution: Reversing the chain rule. If you see a function and its derivative multiplied together inside an integral, substitute a variable (like \(u\)) to simplify the expression.
• Integration by Parts: Reversing the product rule for differentiation. Formula: \(\int u \, dv = uv - \int v \, du\). (Tip: Use the LIATE rule to choose \(u\): Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential).
• Partial Fractions: Breaking down complex rational fractions (like \(\frac{1}{(x-1)(x+2)}\)) into simpler individual fractions that are easy to integrate.

Examples

• \(\int 2x \cos(x^2) dx\). Let \(u = x^2\), then \(du = 2x dx\). The integral becomes \(\int \cos(u) du = \sin(u) + C = \sin(x^2) + C\).

Real-Life Use

Pharmacokinetics uses partial fraction decomposition to model how drugs are absorbed and eliminated by human organs over time.

Meaning

Using definite integrals to find physical areas enclosed by curves, straight lines, or intersecting functions.

Explanation

• Area under a single curve \(y = f(x)\) from \(x = a\) to \(x = b\): $$\text{Area} = \int_{a}^{b} f(x) dx$$
• Area between two intersecting curves \(f(x)\) and \(g(x)\): $$\text{Area} = \int_{a}^{b} [f(x) - g(x)] dx$$ (where \(f(x)\) is the upper curve and \(g(x)\) is the lower curve).

Examples

• Finding the area between \(y = x\) and \(y = x^2\) from \(x = 0\) to \(x = 1\): $$\int_{0}^{1} (x - x^2) dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$$

Real-Life Use

Architects and land surveyors use definite integrals to calculate the exact surface area of irregularly shaped plots of land, lakes, or structural roofing components to estimate material costs.

Meaning

A differential equation is an equation that contains an unknown function and one or more of its derivatives. Instead of solving for a number \(x\), you are solving for an entire function \(y(x)\).

Definitions

1. Order: The highest derivative present in the differential equation.
2. Degree: The highest power of the highest-order derivative (after the equation is made polynomial in its derivatives).

Explanation

• Separable Equations: Move all \(y\) terms with \(dy\) and all \(x\) terms with \(dx\) to opposite sides, then integrate both sides.
• Homogeneous Equations: Differential equations where every term can be written as a function of the ratio \(\frac{y}{x}\). Solved using the substitution \(y = vx\).
• Linear Differential Equations: Equations of the form \(\frac{dy}{dx} + P(x)y = Q(x)\). Solved using an Integrating Factor (IF) given by \(I.F. = e^{\int P(x)dx}\).

Examples

• \(\frac{dy}{dx} = \frac{y}{x}\) is a first-order, first-degree differential equation. Separating variables gives \(\frac{dy}{y} = \frac{dx}{x}\), which integrates to \(\ln y = \ln x + C\).

Real-Life Use

Differential equations rule the universe! Newton's Law of Cooling, population growth models, electrical circuit analysis (RLC circuits), and weather prediction models are all written entirely as differential equations.

Problem 1: Find the intervals in which the function \(f(x) = 2x^3 - 9x^2 + 12x + 15\) is strictly increasing or strictly decreasing.

Solution:

• Step 1: Find the first derivative \(f'(x)\). $$f'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x + 15) = 6x^2 - 18x + 12$$
• Step 2: Factor the derivative. $$f'(x) = 6(x^2 - 3x + 2) = 6(x - 1)(x - 2)$$
• Step 3: Find critical points by setting \(f'(x) = 0\). $$6(x - 1)(x - 2) = 0 \implies x = 1 \text{ and } x = 2$$
• Step 4: Test intervals on the number line: \((-\infty, 1)\), \((1, 2)\), and \((2, \infty)\).
  • For \(x = 0\) (in \((-\infty, 1)\)): \(f'(0) = 6(0-1)(0-2) = +12 > 0\) (Increasing).
  • For \(x = 1.5\) (in \((1, 2)\)): \(f'(1.5) = 6(1.5-1)(1.5-2) = 6(0.5)(-0.5) = -1.5 < 0\) (Decreasing).
  • For \(x = 3\) (in \((2, \infty)\)): \(f'(3) = 6(3-1)(3-2) = +12 > 0\) (Increasing).
• Answer: The function is strictly increasing on \((-\infty, 1) \cup (2, \infty)\) and strictly decreasing on \((1, 2)\).

Problem 2: Solve the linear differential equation \(\frac{dy}{dx} + \frac{y}{x} = x^2\).

Solution:

• Step 1: Identify \(P(x)\) and \(Q(x)\) from the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\). Here, \(P(x) = \frac{1}{x}\) and \(Q(x) = x^2\).
• Step 2: Calculate the Integrating Factor (IF). $$I.F. = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$$
• Step 3: Write the general solution formula: \(y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C\). $$y \cdot x = \int x^2 \cdot x \, dx + C$$ $$xy = \int x^3 dx + C$$
• Step 4: Integrate the right side. $$xy = \frac{x^4}{4} + C$$
• Answer: \(y = \frac{x^3}{4} + \frac{C}{x}\).

When working in professional engineering, data science, and quantitative finance, modern software tools handle heavy calculus calculations instantly:

• Python (Libraries like SymPy, SciPy, and NumPy): Used by data scientists to run symbolic differentiation, solve complex differential equations, and compute multi-variable integrals.
• MATLAB / Simulink: The industry standard for aerospace, automotive, and robotics engineering to model dynamic systems governed by differential equations.
• Wolfram Mathematica / GeoGebra: Excellent educational and professional tools for visualizing 3D surfaces, tangent planes, and calculating definite integrals graphically.

Project Title: Optimizing Suspense and Energy Efficiency in Roller Coaster Design

Background & Scenario:
Imagine you are the lead structural and mechanical engineer for an elite amusement park design firm. Your team has been tasked with designing the signature drop and looping section of a brand-new roller coaster named "The Calculus Thrill." Safety regulations mandate that the track must be completely smooth (continuous and differentiable everywhere) to prevent sudden jolts that could injure riders. Furthermore, the park executives want to maximize the maximum downward acceleration to give riders an unforgettable thrill, while keeping the total steel track material usage minimized to stay within budget.

Engineering Analysis Using Calculus:

1. Continuity & Differentiability Check: The transition ramp connecting the flat station platform to the steep drop is modeled by a piecewise function. Engineers use limits to ensure the transition point connects seamlessly without any gaps (Continuity). They verify that the first derivative of the track curve matches perfectly at the joining points so there are no sharp corners (Differentiability).
2. Derivatives for Velocity and Acceleration: The height \(h(t)\) of the coaster car as a function of time \(t\) seconds is given by \(h(t) = -4.9t^2 + 20t + 50\).
   • Taking the first derivative gives the vertical velocity: \(v(t) = h'(t) = -9.8t + 20\).
   • Taking the second derivative gives the gravitational acceleration experienced by the riders: \(a(t) = h''(t) = -9.8 \text{ meters per second squared}\).
3. Optimization (Maxima/Minima): To find the exact peak height and maximum speed during the descent, the engineering team sets the first derivative of the velocity equation to zero, using critical point analysis to identify optimal structural turning points.
4. Integration for Material Costs: To order the exact amount of forged steel tubing required for the curved dips, the team uses definite integrals. By calculating the arc length and area under the track curve from \(x = 0\) meters to \(x = 100\) meters, they compute the precise surface area and material volume needed. This saves the company hundreds of thousands of dollars in surplus material waste.
5. Differential Equations in Action: Atmospheric air resistance and friction damping are modeled using a first-order linear differential equation: \(\frac{dv}{dt} + kv = g\), ensuring that terminal velocity calculations remain within safe international amusement park safety standards.

Through this comprehensive application of calculus, the roller coaster is successfully built—delivering maximum excitement while maintaining 100 percent structural safety.

1. Q1. What is the value of \(\lim_{x \to 0} \frac{\sin x}{x}\)?
   A) \(0\)
   B) \(1\)
   C) Undefined
   D) \(-1\)
   Answer: B
   Explanation: This is a standard trigonometric limit which evaluates to \(1\) (proven using L'Hopital's Rule or geometric proofs).


2. Q2. Which of the following functions is continuous everywhere but NOT differentiable at \(x = 0\)?
   A) \(f(x) = x^2\)
   B) \(f(x) = |x|\)
   C) \(f(x) = \frac{1}{x}\)
   D) \(f(x) = \ln(x)\)
   Answer: B
   Explanation: The absolute value function \(f(x) = |x|\) forms a sharp "V" vertex at the origin, meaning it has a continuous graph but lacks a unique tangent line at \(x = 0\).


3. Q3. What is the derivative of \(y = e^{3x}\) with respect to \(x\)?
   A) \(e^{3x}\)
   B) \(3e^{3x}\)
   C) \(\frac{1}{3}e^{3x}\)
   D) \(3xe^{3x-1}\)
   Answer: B
   Explanation: Using the chain rule, the derivative of \(e^{u}\) is \(e^{u} \cdot \frac{du}{dx}\). Here \(\frac{d}{dx}(3x) = 3\), giving \(3e^{3x}\).


4. Q4. If \(x = \cos t\) and \(y = \sin t\), what is \(\frac{dy}{dx}\)?
   A) \(-\tan t\)
   B) \(\cot t\)
   C) \(-\cot t\)
   D) \(\tan t\)
   Answer: C
   Explanation: \(\frac{dy}{dt} = \cos t\) and \(\frac{dx}{dt} = -\sin t\). Therefore, \(\frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t\).


5. Q5. What does the second derivative \(f''(x)\) represent geometrically?
   A) Slope of the curve
   B) Concavity and inflection points
   C) Total area under the curve
   D) Y-intercept
   Answer: B
   Explanation: The second derivative determines whether a curve is concave up (\(f''(x) > 0\)) or concave down (\(f''(x) < 0\)), and helps locate inflection points.


6. Q6. Rolle's Theorem requires which of the following conditions for \(f(x)\) on interval \([a, b]\)?
   A) \(f(a) \neq f(b)\)
   B) \(f(a) = f(b)\)
   C) \(f'(x) = 0\) everywhere
   D) \(f(x)\) must be discontinuous
   Answer: B
   Explanation: One of the strict requirements of Rolle's Theorem is that the function values at the endpoints must be equal (\(f(a) = f(b)\)).


7. Q7. What is the anti-derivative (integral) of \(\frac{1}{x}\) with respect to \(x\)?
   A) \(-\frac{1}{x^2} + C\)
   B) \(\ln|x| + C\)
   C) \(e^x + C\)
   D) \(x \ln x + C\)
   Answer: B
   Explanation: The fundamental integration rule states that \(\int \frac{1}{x} dx = \ln|x| + C\).


8. Q8. Which integration technique is best suited for evaluating \(\int x e^x dx\)?
   A) Substitution
   B) Partial Fractions
   C) Integration by Parts
   D) Trigonometric Substitution
   Answer: C
   Explanation: The integral is a product of an algebraic function (\(x\)) and an exponential function (\(e^x\)), making Integration by Parts (\(\int u dv = uv - \int v du\)) the ideal method.


9. Q9. What is the order of the differential equation \(\left(\frac{d^2y}{dx^2}\right)^3 + 4\left(\frac{dy}{dx}\right) = y\)?
   A) \(1\)
   B) \(2\)
   C) \(3\)
   D) \(6\)
   Answer: B
   Explanation: The order of a differential equation is the order of the highest derivative present. Here, the highest derivative is the second derivative (\(\frac{d^2y}{dx^2}\)), so the order is \(2\).


10. Q10. What is the Integrating Factor (IF) for the linear differential equation \(\frac{dy}{dx} + 2y = \cos x\)?
    A) \(e^{2x}\)
    B) \(2x\)
    C) \(e^x\)
    D) \(\sin x\)
    Answer: A
    Explanation: Here \(P(x) = 2\). The integrating factor is given by \(I.F. = e^{\int P(x) dx} = e^{\int 2 dx} = e^{2x}\).

1. Q1. Examine the continuity of the function \(f(x) = \begin{cases} x^2 + 1, & \text{if } x \le 1 \\ 2x, & \text{if } x > 1 \end{cases}\) at \(x = 1\).
   Answer:
   - Left-Hand Limit (\(\lim_{x \to 1^-} f(x)\)) \(= 1^2 + 1 = 2\).
   - Right-Hand Limit (\(\lim_{x \to 1^+} f(x)\)) \(= 2(1) = 2\).
   - Value of function at \(x = 1\): \(f(1) = 1^2 + 1 = 2\).
   - Since Left-Hand Limit = Right-Hand Limit = Function Value (\(2 = 2 = 2\)), the function is continuous at \(x = 1\).


2. Q2. Find \(\frac{dy}{dx}\) if \(x^3 + y^3 = 3axy\) (Implicit Differentiation).
   Answer:
   Differentiate both sides with respect to \(x\): \(3x^2 + 3y^2\frac{dy}{dx} = 3a\left(x\frac{dy}{dx} + y\cdot 1\right)\)
   Divide by \(3\): \(x^2 + y^2\frac{dy}{dx} = ax\frac{dy}{dx} + ay\)
   Group \(\frac{dy}{dx}\) terms on one side: \(\frac{dy}{dx}(y^2 - ax) = ay - x^2\)
   \(\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}\)


3. Q3. Find the derivative of \(y = x^{\sin x}\) using logarithmic differentiation.
   Answer:
   Take \(\ln\) on both sides: \(\ln y = \sin x \cdot \ln x\)
   Differentiate with respect to \(x\): \(\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}\)
   Multiply by \(y\): \(\frac{dy}{dx} = x^{\sin x} \left( \cos x \ln x + \frac{\sin x}{x} \right)\)


4. Q4. Verify Rolle's Theorem for \(f(x) = x^2 - 4x + 3\) on the interval \([1, 3]\).
   Answer:
   - \(f(x)\) is a polynomial, so it is continuous on \([1, 3]\) and differentiable on \((1, 3)\).
   - Check endpoint values: \(f(1) = 1^2 - 4(1) + 3 = 0\), and \(f(3) = 3^2 - 4(3) + 3 = 0\). Since \(f(1) = f(3)\), Rolle's conditions are met.
   - Find \(f'(x) = 2x - 4\).
   - Set \(f'(c) = 0 \implies 2c - 4 = 0 \implies c = 2\).
   - Since \(c = 2\) lies inside \((1, 3)\), Rolle's Theorem is verified.


5. Q5. Evaluate the definite integral \(\int_{0}^{\pi/2} \cos^2 x dx\).
   Answer:
   Use the trigonometric identity \(\cos^2 x = \frac{1 + \cos(2x)}{2}\):
   \(\int_{0}^{\pi/2} \frac{1 + \cos(2x)}{2} dx = \frac{1}{2} \left[ x + \frac{\sin(2x)}{2} \right]_0^{\pi/2}\)
   \(= \frac{1}{2} \left[ \left(\frac{\pi}{2} + \frac{\sin(\pi)}{2}\right) - (0 + 0) \right] = \frac{1}{2} \left(\frac{\pi}{2}\right) = \frac{\pi}{4}\)


6. Q6. Evaluate \(\int x^2 e^x dx\) using Integration by Parts.
   Answer:
   Using \(\int u dv = uv - \int v du\), let \(u = x^2\) and \(dv = e^x dx\) (so \(du = 2x dx\) and \(v = e^x\)).
   \(= x^2 e^x - \int 2x e^x dx\)
   Apply parts again on \(\int x e^x dx\) (let \(u = x, dv = e^x dx\)):
   \(= x^2 e^x - 2 \left( x e^x - \int e^x dx \right)\)
   \(= x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C\)


7. Q7. Find the area of the region bounded by the parabola \(y^2 = 4x\) and the line \(y = 2x\).
   Answer:
   Find intersection points by substituting \(y = 2x\) into \(y^2 = 4x\):
   \((2x)^2 = 4x \implies 4x^2 - 4x = 0 \implies 4x(x - 1) = 0 \implies x = 0, x = 1\).
   Area formula between curves:
   \(\int_{0}^{1} (\sqrt{4x} - 2x) dx = \int_{0}^{1} (2\sqrt{x} - 2x) dx\)
   \(= \left[ 2\left(\frac{x^{3/2}}{3/2}\right) - x^2 \right]_0^1 = \left[ \frac{4}{3}x^{3/2} - x^2 \right]_0^1 = \frac{4}{3} - 1 = \frac{1}{3} \text{ sq. units}\)


8. Q8. Find the general solution of the differential equation \(\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}\) (Separation of Variables).
   Answer:
   Separate variables: \(\frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}\)
   Integrate both sides: \(\tan^{-1}(y) = \tan^{-1}(x) + C\)


9. Q9. Find the order and degree of the differential equation \(\frac{d^2y}{dx^2} + 3\left(\frac{dy}{dx}\right)^2 = \sin x\).
   Answer:
   - Order: \(2\) (since the highest derivative is the second derivative \(\frac{d^2y}{dx^2}\)).
   - Degree: \(1\) (since the exponent of the highest-order derivative term is \(1\)).


10. Q10. A ladder \(5\) meters long is leaning against a wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of \(2\) meters per second. How fast is its height on the wall decreasing when the foot of the ladder is \(4\) meters away from the wall?
    Answer:
    - Let \(x\) be the distance of the bottom from the wall, and \(y\) be the height on the wall.
    - By Pythagoras theorem: \(x^2 + y^2 = 5^2 = 25\). Given \(\frac{dx}{dt} = 2\) m/s, and when \(x = 4\), \(4^2 + y^2 = 25 \implies y = 3\) meters.
    - Differentiate with respect to time \(t\):
    \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies x\frac{dx}{dt} + y\frac{dy}{dt} = 0\)
    - Substitute values:
    \((4)(2) + (3)\frac{dy}{dt} = 0 \implies 8 + 3\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{8}{3} \text{ m/s}\)
    - Answer: The height on the wall is decreasing at a rate of \(\frac{8}{3}\) meters per second.

• Mnemonic for Integration by Parts (LIATE): When choosing \(u\), follow the order: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential.
• Frequently Confused Concept 1: Derivative vs. Integral of \(e^x\). Remember: \(\frac{d}{dx}(e^x) = e^x\) and \(\int e^x dx = e^x + C\). Exponential functions are unique because they are their own derivatives and integrals!
• Frequently Confused Concept 2: Continuity vs. Differentiability. Always remember: A function can be continuous at a corner point without being differentiable (like \(|x|\) at zero). Differentiability is stricter; continuity is more relaxed.

Authority Citations:
- Newton, Sir Isaac. Philosophiae Naturalis Principia Mathematica (Foundational framework for fluxions and differential calculus).
- Leibniz, Gottfried Wilhelm. Nova Methodus pro Maximis et Minimis (Historical introduction of differential notation \(dx\) and \(dy\)).