Introduction To Mathematical Induction

 

What is Mathematical Induction?

Think of mathematical induction like knocking over a row of standing dominoes. Instead of knocking each one down individually, you only need to ensure two things:

1. The first domino falls.


2. If any given domino falls, it will inevitably knock over the next one.

In mathematics, this is a powerful technique used to prove that a statement, formula, or property is true for all positive integers (n = 1, 2, 3, ...).

A formal proof by induction consists of two essential steps:

• The Base Case: Prove that the statement holds true for the very first value, usually n = 1.


• The Inductive Step: Prove that if the statement holds true for an arbitrary integer k (this assumption is called the Inductive Hypothesis), then it must also be true for the next integer k + 1.

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Two Classic Examples

Example 1: Sum of the First n Positive Integers

We want to prove the formula: 1 + 2 + 3 + ... + n = [n(n + 1)] / 2

• Base Case (n = 1): Left Side = 1. Right Side = [1(1 + 1)] / 2 = 2 / 2 = 1. Both sides match, so the base case holds.


• Inductive Step: Assume it works for n = k, meaning: 1 + 2 + ... + k = [k(k + 1)] / 2. Now, we must prove it works for n = k + 1. Add (k + 1) to both sides of our assumption, factor it out, and it simplifies perfectly to: [(k + 1)(k + 2)] / 2. The proof is complete.

Example 2: Divisibility

We want to prove that (n³ - n) is always divisible by 3 for any positive integer n.

• Base Case (n = 1): 1³ - 1 = 0. Since 0 is divisible by 3 (0 = 3 x 0), the base case holds.


• Inductive Step: Assume (k³ - k) is divisible by 3, so k³ - k = 3m. Now check for n = k + 1: (k + 1)³ - (k + 1). Expanding and rearranging gives us (k³ - k) + 3k² + 3k. Substituting our assumption gives 3m + 3k² + 3k, which factors out to 3(m + k² + k). Because the expression can be factored by 3, it is guaranteed to be divisible by 3.

Solving Problems On Differentiation Of Trigonometric Functions

 

Rules of Differentiation for Trigonometric Functions

A comprehensive guide with formulas, objective questions, and step-by-step solutions.

1. Derivative of Sine [d/dx(sin x)]

The derivative of the sine function is the cosine function.

d/dx(sin x) = cos x

Objective Question 1: Find the derivative of f(x) = sin(3x).

• A) cos(3x)
• B) 3cos(3x)
• C) -3cos(3x)
• D) 3sin(3x)

Correct Answer: B Solution:
Using the Chain Rule:
Let u = 3x, so du/dx = 3.
d/dx(sin u) = cos u * du/dx
d/dx(sin(3x)) = cos(3x) * 3 = 3cos(3x)

2. Derivative of Cosine [d/dx(cos x)]

The derivative of the cosine function is the negative sine function.

d/dx(cos x) = -sin x

Objective Question 2: Differentiate y = cos(x^2) with respect to x.

• A) -sin(x^2)
• B) 2x+sin(x^2)
• C) -2x*sin(x^2)
• D) -2x*cos(x^2)

Correct Answer: C Solution:
Apply the Chain Rule:
The derivative of the outer function cos(u) is -sin(u).
The derivative of the inner function x^2 is 2x.
Multiplying them together gives: -2x*sin(x^2).

3. Derivative of Tangent [d/dx(tan x)]

The derivative of the tangent function is the secant squared function.

d/dx(tan x) = sec^2 x

Objective Question 3: What is d/dx(5tan x)?

• A) 5sec^2 x
• B) sec^2(5x)
• C) 5cot x
• D) -5sec^2 x

Correct Answer: A Solution:
By the constant multiple rule, pull out the constant 5:
d/dx(5tan x) = 5 * d/dx(tan x) = 5sec^2 x.

4. Derivative of Cosecant [d/dx(csc x)]

The derivative of the cosecant function is the negative product of cosecant and cotangent.

d/dx(csc x) = -csc x * cot x

Objective Question 4: Find the derivative of f(x) = csc(4x).

• A) -csc(4x)cot(4x)
• B) 4csc(4x)cot(4x)
• C) -4csc(4x)cot(4x)
• D) -4sec(4x)tan(4x)

Correct Answer: C Solution:
Using the Chain Rule:
Differentiate the outer function: -csc(4x)cot(4x).
Multiply by the derivative of the inner function (4x), which is 4.
Result: -4csc(4x)cot(4x).

5. Derivative of Secant [d/dx(sec x)]

The derivative of the secant function is the product of secant and tangent.

d/dx(sec x) = sec x * tan x

Objective Question 5: Evaluate the derivative of y = x*sec x.

• A) sec x * tan x
• B) sec x * (1 + x*tan x)
• C) sec x + tan x
• D) x*sec x*tan x

Correct Answer: B Solution:
Apply the Product Rule [d/dx(uv) = u'v + uv']:
Let u = x implies u' = 1.
Let v = sec x implies v' = sec x * tan x.
dy/dx = (1)(sec x) + (x)(sec x * tan x) = sec x + x*sec x*tan x.
Factoring out sec x gives: sec x * (1 + x*tan x).

6. Derivative of Cotangent [d/dx(cot x)]

The derivative of the cotangent function is the negative cosecant squared function.

d/dx(cot x) = -csc^2 x

Objective Question 6: If y = cot x, find dy/dx at x = pi/4.

• A) -1
• B) -2
• C) 1
• D) 2

Correct Answer: B Solution:
First, find the general derivative: dy/dx = -csc^2 x.
Recall that csc x = 1/sin x. At x = pi/4, sin(pi/4) = 1/sqrt(2).
Therefore, csc(pi/4) = sqrt(2).
Squaring it gives [sqrt(2)]^2 = 2.
Substituting this back into our formula yields: -2.

Solving Calculus Problems On Integration

 

Mastering Integration

Definition, 10 Core Rules, Practice Questions, and Detailed Solutions

What is Integration?

Integration is a foundational concept in calculus that serves as the inverse operation to differentiation. While differentiation determines the rate of change at a given point, integration accumulates these values to calculate total sizes, most notably the net area bounded under a curve.

• Indefinite Integrals: Represent families of functions and always include an arbitrary constant (C).
• Definite Integrals: Evaluated across specific physical intervals to provide a concrete numeric value.

1. The Power Rule

Used to integrate variable bases raised to fixed numerical exponents.

∫ xⁿ dx = (xⁿ⁺¹ / (n + 1)) + C  (where n ≠ -1)

Question 1.1: Find the indefinite integral: ∫ x⁴ dx.

• A) 4x³ + C
• B) (1/5)x⁵ + C
• C) 5x⁵ + C
• D) x⁵ + C

Correct Answer: B

Solution: Raise power by 1 and divide by the new exponent: (4+1=5). Result = (1/5)x⁵ + C.

Question 1.2: Evaluate: ∫ 1/x³ dx.

• A) -1/(2x²) + C
• B) 1/(4x⁴) + C
• C) -3/x⁴ + C
• D) log(x³) + C

Correct Answer: A

Solution: Rewrite as negative exponent: ∫ x^-3 dx. Apply rule: x^-2 / -2 + C = -1/(2x²) + C.

2. Constant & Constant Multiple Rule

Constants can be factored directly out of integration operations cleanly.

∫ k dx = kx + C  |  ∫ k·f(x) dx = k ∫ f(x) dx

Question 2.1: Find ∫ 7 dx.

• A) 7 + C
• B) 0 + C
• C) 7x + C
• D) 7/2 x² + C

Correct Answer: C

Solution: The integral of any standalone constant k with respect to x is always kx + C. Thus, 7x + C.

Question 2.2: Evaluate ∫ 6x² dx.

• A) 2x³ + C
• B) 18x³ + C
• C) 6x³ + C
• D) 3x² + C

Correct Answer: A

Solution: Factor out 6: 6 ∫ x² dx = 6 · (x³ / 3) = 2x³ + C.

3. Sum and Difference Rule

Integrals can be distributed cleanly across independent additive or subtractive elements.

∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx

Question 3.1: Evaluate ∫ (3x² + 2x) dx.

• A) 6x + 2 + C
• B) x³ + x² + C
• C) 3x³ + 2x² + C
• D) x³ + 2x + C

Correct Answer: B

Solution: Split terms: ∫ 3x² dx + ∫ 2x dx = x³ + x² + C. Result = x³ + x² + C.

Question 3.2: Find ∫ (5 - x) dx.

• A) 5x - (1/2)x² + C
• B) 5x - x + C
• C) -1 + C
• D) 5x - x² + C

Correct Answer: A

Solution: Integrating independent blocks gives: ∫ 5 dx - ∫ x dx = 5x - (1/2)x² + C.

4. The Exponential Rule

Natural exponential operations are unique as they serve as their own primary integrals.

∫ e^x dx = e^x + C  |  ∫ e^ax dx = (1/a)e^ax + C

Question 4.1: Evaluate ∫ 4e^x dx.

• A) e^x + C
• B) 4e^x + C
• C) 2e^2x + C
• D) 4xe^x-1 + C

Correct Answer: B

Solution: Pull out the constant scale factor directly: 4 ∫ e^x dx = 4e^x + C.

Question 4.2: Find ∫ e^5x dx.

• A) 5e^5x + C
• B) e^5x + C
• C) (1/5)e^5x + C
• D) e^6x/6 + C

Correct Answer: C

Solution: Divide structural expression by the internal derivative coefficient (5): (1/5)e^5x + C.

5. The Reciprocal Rule (Logarithmic Rule)

The solution when evaluating functions with a power profile of exactly -1.

∫ (1/x) dx = ln|x| + C

Question 5.1: Find the integral: ∫ 3/x dx.

• A) -3/x² + C
• B) 3 ln|x| + C
• C) ln|3x| + C
• D) 3x⁰ + C

Correct Answer: B

Solution: Separate the numerator value: 3 ∫ (1/x) dx = 3 ln|x| + C.

Question 5.2: Evaluate ∫ 1/(4x) dx.

• A) (1/4) ln|x| + C
• B) ln|4x| + C
• C) 4 ln|x| + C
• D) -1/(16x²) + C

Correct Answer: A

Solution: Extract scalar fractions cleanly before processing: (1/4) ∫ (1/x) dx = (1/4) ln|x| + C.

6. Trigonometric Rules (Sine and Cosine)

Core cyclic rules tracking foundational derivatives in inverse format.

∫ sin(x) dx = -cos(x) + C  |  ∫ cos(x) dx = sin(x) + C

Question 6.1: Evaluate ∫ -2 cos(x) dx.

• A) 2 sin(x) + C
• B) -2 sin(x) + C
• C) -2 cos(x) + C
• D) 2 cos(x) + C

Correct Answer: B

Solution: The integral of cos(x) is sin(x). Multiplying by -2 yields -2 sin(x) + C.

Question 6.2: Find ∫ (x - sin(x)) dx.

• A) (1/2)x² - cos(x) + C
• B) x² + cos(x) + C
• C) (1/2)x² + cos(x) + C
• D) 1 - cos(x) + C

Correct Answer: C

Solution: ∫ x dx - ∫ sin(x) dx = (1/2)x² - (-cos(x)) + C = (1/2)x² + cos(x) + C.

7. Advanced Trigonometric Rules (Secant Squared)

Since the derivative of tangent functions equals secant squared, integration reverses this track perfectly.

∫ sec²(x) dx = tan(x) + C

Question 7.1: Evaluate ∫ 5 sec²(x) dx.

• A) 5 tan(x) + C
• B) -5 tan(x) + C
• C) 5 sec(x) + C
• D) 5/3 sec³(x) + C

Correct Answer: A

Solution: Factor out the scale parameter: 5 ∫ sec²(x) dx = 5 tan(x) + C.

Question 7.2: Find ∫ sec²(2x) dx.

• A) tan(2x) + C
• B) 2 tan(2x) + C
• C) (1/2) tan(2x) + C
• D) sec(2x) tan(2x) + C

Correct Answer: C

Solution: Divide by the internal linear derivative coefficient (2) to scale back the transformation: (1/2) tan(2x) + C.

8. Integration by Substitution (u-Substitution)

The inverse version of the Chain Rule. Used when an expression contains both an inner function and its derivative.

∫ f(g(x))·g'(x) dx = ∫ f(u) du, where u = g(x)

Question 8.1: Evaluate ∫ 2x e^x2 dx.

• A) e^x2 + C
• B) 2e^x2 + C
• C) x²e^x + C
• D) e^x + C

Correct Answer: A

Solution: Set u = x², giving du = 2x dx. Substituting gives ∫ e^u du = e^u + C = e^x2 + C.

Question 8.2: Find ∫ (2x + 3)⁵ dx.

• A) (2x + 3)⁶ / 6 + C
• B) (2x + 3)⁶ / 12 + C
• C) 2(2x + 3)⁶ + C

Solving Problems On Application of Differentiation

 

Applications of Differentiation: Core Concepts

Differentiation is not just an abstract algebraic exercise; it is a powerful tool used to analyze behavior, rates, and limits in the real world. Here are the three most common applications found in calculus:

• 1. Rates of Change: Since a derivative measures how fast a dependent variable changes relative to an independent variable, it represents real-world rates. For example, if an object's position is given by a function of time, its velocity is the first derivative, and its acceleration is the second derivative.
• 2. Turning Points (Maxima and Minima): To find where a curve reaches its highest or lowest point, we look for where the slope of the tangent line equals zero (^dy⁄_dx = 0). These are called stationary points.
• 3. Optimization: This is the business and engineering application of finding turning points. It involves constructing a mathematical model to calculate maximum possible profit, minimum material waste, or optimal structural dimensions under specific constraints.

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Guided Examples

Example 1 (Kinematics / Rate of Change):
The distance s (in meters) traveled by a particle in time t (in seconds) is given by the equation:
s = 2t³ - 5t + 3
Find the velocity of the particle at t = 3 seconds.

Solution:
Velocity (v) is defined as the instantaneous rate of change of distance with respect to time: v = ^ds⁄_dt.
1. Differentiate the equation with respect to t using the Power Rule:
v = ^ds⁄_dt = 6t² - 5
2. Substitute t = 3 into the velocity derivative:
v = 6(3)² - 5 = 6(9) - 5 = 54 - 5 = 49 m/s.
The velocity at 3 seconds is 49 m/s.

Example 2 (Stationary Points):
Find the coordinates of the turning point for the curve:
y = x² - 4x + 7

Solution:
1. Turning points occur exclusively where the gradient/slope is zero (^dy⁄_dx = 0). Differentiate the function:
^dy⁄_dx = 2x - 4
2. Set the derivative equal to zero and isolate x:
2x - 4 = 0 → 2x = 4 → x = 2
3. Find the corresponding y value by substituting x = 2 back into the original curve equation:
y = (2)² - 4(2) + 7 = 4 - 8 + 7 = 3.
The stationary turning point coordinates are (2, 3).

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Application of Differentiation Practice Quiz

Question 1: The displacement of an object is given by s = t² - 6t + 8. At what time t does the object momentarily come to rest (velocity = 0)?

• A) t = 2 seconds
• B) t = 3 seconds
• C) t = 6 seconds
• D) t = 0 seconds

View Answer & mSolution

Correct Answer: B

Solution: Velocity is the derivative of displacement: v = ^ds⁄_dt = 2t - 6.
The term "at rest" means velocity equals 0. Set the expression to 0:
2t - 6 = 0 → 2t = 6 → t = 3 seconds.

Question 2: Find the x-coordinate of the maximum turning point for the curve y = -x² + 8x - 12.

• A) x = 4
• B) x = -4
• C) x = 8
• D) x = 2

View Answer & Solution

Correct Answer: A

Solution: Differentiate the function to find the gradient expression:
^dy⁄_dx = -2x + 8
Set the derivative to zero for a stationary point:
-2x + 8 = 0 → 2x = 8 → x = 4.

Question 3: A profit function for selling x items is given by P(x) = 400x - 2x². How many items must be sold to maximize profit?

• A) 400 items
• B) 200 items
• C) 100 items
• D) 50 items

View Answer & Solution

Correct Answer: C

Solution: To find maximum profit, find the marginal profit derivative P'(x) and set it to 0:
P'(x) = 400 - 4x
Set equal to zero:
400 - 4x = 0 → 4x = 400 → x = 100 items.

Question 4: The velocity of a moving body is given by v = 3t² + 2t. Find its acceleration at t = 2 seconds.

• A) 14 m/s²
• B) 16 m/s²
• C) 12 m/s²
• D) 8 m/s²

View Answer & Solution

Correct Answer: A

Solution: Acceleration (a) is the derivative of velocity over time: a = ^dv⁄_dt.
Differentiate the velocity function:
a = 6t + 2
Substitute t = 2:
a = 6(2) + 2 = 12 + 2 = 14 m/s².

Question 5: If a curve is given by y = x³ - 3x, what are the x-values of its two stationary points?

• A) x = 0 and x = 3
• B) x = 1 and x = -1
• C) x = 0 and x = 1
• D) x = 3 and x = -3

View Answer & Solution

Correct Answer: B

Solution: Differentiate the cubic equation to find its slope formula:
^dy⁄_dx = 3x² - 3
Set the expression equal to 0:
3x² - 3 = 0 → 3x² = 3 → x² = 1
Taking the square root gives: x = 1 and x = -1.

Solving Calculus Problems On Implicit Differentiation

 

Understanding Implicit Differentiation

In calculus, we usually work with explicit functions, where one variable is isolated directly on one side of the equation—for example, y = x² + 3x. Differentiating these is straightforward because y is stated explicitly in terms of x.

However, many equations mix x and y together in a way that is difficult or impossible to solve directly for y. These are called implicit functions (such as x² + y² = 25 or x³ + y³ = 3xy).

To find ^dy⁄_dx for these expressions, we use Implicit Differentiation. The process follows three core steps:

1. Differentiate both sides of the equation with respect to x.
2. Treat y as a function of x. This means every time you differentiate a term containing y, you must apply the Chain Rule and multiply that term by ^dy⁄_dx.
3. Use algebra to isolate and solve for ^dy⁄_dx.

Guided Example

Problem: Find ^dy⁄_dx for the curve x² + y² = 36.

Step 1: Differentiate term by term with respect to x:
The derivative of x² is 2x.
The derivative of y² (using the Chain Rule) is 2y · ^dy⁄_dx.
The derivative of the constant 36 is 0.

Putting it together:
2x + 2y(^dy⁄_dx) = 0

Step 2: Isolate the ^dy⁄_dx term:
Subtract 2x from both sides:
2y(^dy⁄_dx) = -2x

Divide both sides by 2y:
^dy⁄_dx = ^-2x⁄_2y

Simplifying gives the final result:
^dy⁄_dx = -^x⁄_y

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Implicit Differentiation Practice Quiz

Question 1: Find ^dy⁄_dx if y³ - x³ = 8.

• A) ^x2⁄_y²
• B) -^x2⁄_y²
• C) ^y2⁄_x²
• D) 3x²

View Answer & Solution

Correct Answer: A

Solution:
1. Differentiate both sides with respect to x:
3y²(^dy⁄_dx) - 3x² = 0
2. Move the x term to the right side:
3y²(^dy⁄_dx) = 3x²
3. Divide by 3y² to isolate the derivative:
^dy⁄_dx = ^3x2⁄_3y² = ^x2⁄_y²

Question 2: Find the derivative ^dy⁄_dx for the curve expression xy = 4.

• A) ¹⁄_x
• B) -^y⁄_x
• C) -^x⁄_y
• D) 0

View Answer & Solution

Correct Answer: B

Solution:
1. Use the Product Rule on the term xy: (first · derivative of second) + (second · derivative of first).
x(^dy⁄_dx) + y(1) = 0
2. Isolate the derivative term:
x(^dy⁄_dx) = -y
3. Divide by x:
^dy⁄_dx = -^y⁄_x

Question 3: Differentiate implicitly to find ^dy⁄_dx given x + sin(y) = 1.

• A) -cos(y)
• B) ¹⁄_cos(y)
• C) -¹⁄_cos(y)
• D) -sin(y)

View Answer & Solution

Correct Answer: C

Solution:
1. Differentiate term by term with respect to x:
1 + cos(y) · ^dy⁄_dx = 0
2. Move 1 over to the right hand side:
cos(y) · ^dy⁄_dx = -1
3. Divide by cos(y) to finish solving:
^dy⁄_dx = -¹⁄_cos(y)

Question 4: Find ^dy⁄_dx if x² + 3xy = 10.

• A) -^{2x + 3y}⁄_3x
• B) -^2x⁄₃
• C) ^{2x + 3y}⁄_3x
• D) -^{2x + 3}⁄_3x

View Answer & Solution

Correct Answer: A

Solution:
1. Differentiate term by term. Use the product rule for the 3xy part:
2x + [3x(^dy⁄_dx) + 3y(1)] = 0
2x + 3x(^dy⁄_dx) + 3y = 0
2. Keep the ^dy⁄_dx term on the left side, and move the rest to the right:
3x(^dy⁄_dx) = -2x - 3y
3. Factor out the negative sign on the right side and divide by 3x:
3x(^dy⁄_dx) = -(2x + 3y)
^dy⁄_dx = -^{2x + 3y}⁄_3x

Question 5: Find the slope of the tangent line ^dy⁄_dx for the curve y² - x = 5.

• A) 2y
• B) ¹⁄_2y
• C) -¹⁄_2y
• D) 1

View Answer & Solution

Correct Answer: B

Solution:
1. Differentiate implicitly with respect to x:
2y(^dy⁄_dx) - 1 = 0
2. Move the constant -1 to the right side:
2y(^dy⁄_dx) = 1
3. Divide by 2y to isolate the derivative expression:
^dy⁄_dx = ¹⁄_2y

Solving Calculus Problems On Differentiation

 

What is Differentiation?

Differentiation is a fundamental operation in calculus that measures the rate at which a function changes relative to its input variable. Geometrically, finding the derivative of a function at a specific point gives the slope of the tangent line to the curve at that exact point.

If y = f(x), the derivative is denoted as ^dy⁄_dx or f'(x), and it represents the instantaneous rate of change of y with respect to x.

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The First 10 Rules of Differentiation

1. The Constant Rule: The derivative of any constant is zero.
   ^d⁄_dx(c) = 0
2. The Power Rule: Bring the power to the front and subtract one.
   ^d⁄_dx(xⁿ) = n · x^n-1
3. The Constant Multiple Rule: Factor out the constant before differentiating.
   ^d⁄_dx(c · f(x)) = c · f'(x)
4. The Sum Rule: Differentiate terms separately.
   ^d⁄_dx(f(x) + g(x)) = f'(x) + g'(x)
5. The Difference Rule: Differentiate terms separately.
   ^d⁄_dx(f(x) - g(x)) = f'(x) - g'(x)
6. The Product Rule: First times derivative of the second, plus second times derivative of the first.
   ^d⁄_dx(f(x) · g(x)) = f(x)g'(x) + g(x)f'(x)
7. The Quotient Rule: "Low d-high minus high d-low, over the square of what's below."
   ^d⁄_dx( ^f(x)⁄_g(x) ) = g(x)f'(x) - f(x)g'(x) ⁄ _[g(x)]²
8. The Chain Rule: Derivative of the outer function multiplied by the derivative of the inner function.
   ^d⁄_dx(f(g(x))) = f'(g(x)) · g'(x)
9. The Exponential Rule (Base e): The natural exponential function is its own derivative.
   ^d⁄_dx(e^x) = e^x
10. The Logarithmic Rule (Natural Log): The derivative of ln(x) is 1/x.
    ^d⁄_dx(ln(x)) = ¹⁄_x

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Practice Quiz (10 Objective Questions)

Question 1 (Constant Rule): What is the derivative of f(x) = 157?

• A) 157
• B) 1
• C) 0
• D) 157x

View Answer & Solution

Correct Answer: C

Solution: According to the Constant Rule, the rate of change of any constant value is always zero because it does not vary with x. Therefore, ^d⁄_dx(157) = 0.

Question 2 (Power Rule): Find the derivative of f(x) = x⁷.

• A) 7x⁷
• B) 7x⁶
• C) 6x⁷
• D) ^x8⁄₈

View Answer & Solution

Correct Answer: B

Solution: Apply the Power Rule, ^d⁄_dx(xⁿ) = n · x^n-1. Here, n = 7. Bringing the 7 to the front and subtracting 1 from the power gives 7x^7-1 = 7x⁶.

Question 3 (Constant Multiple Rule): Differentiate f(x) = 4x⁵.

• A) 20x⁴
• B) 4x⁴
• C) 20x⁵
• D) 5x⁴

View Answer & Solution

Correct Answer: A

Solution: Keep the constant coefficient 4 aside and differentiate x⁵ using the Power Rule (5x⁴). Then multiply them back together: 4 · (5x⁴) = 20x⁴.

Question 4 (Sum Rule): Find ^dy⁄_dx if y = x³ + x².

• A) 5x⁴
• B) 3x² + x
• C) 3x² + 2x
• D) 6x

View Answer & Solution

Correct Answer: C

Solution: The Sum Rule states you can differentiate each term independently. The derivative of x³ is 3x² and the derivative of x² is 2x. Adding them together yields 3x² + 2x.

Question 5 (Difference Rule): Find the derivative of f(x) = 6x² - 2x.

• A) 12x - 2
• B) 12x
• C) 6x - 2
• D) 12x² - 2

View Answer & Solution

Correct Answer: A

Solution: Differentiate each term independently across the subtraction sign. ^d⁄_dx(6x²) = 12x and ^d⁄_dx(2x) = 2. This leaves us with 12x - 2.

Question 6 (Product Rule): Differentiate y = x² · e^x.

• A) 2x · e^x
• B) x² · e^x + 2x · e^x
• C) x² + e^x
• D) 2x² · e^x

View Answer & Solution

Correct Answer: B

Solution: Let f(x) = x² → f'(x) = 2x, and g(x) = e^x → g'(x) = e^x. Applying the Product Rule f(x)g'(x) + g(x)f'(x), we obtain (x²)(e^x) + (e^x)(2x), which is x²e^x + 2xe^x.

Question 7 (Quotient Rule): Find the derivative of f(x) = ln(x) ⁄ x.

• A) 1 ⁄ x²
• B) (1 - ln(x)) ⁄ x
• C) (1 - ln(x)) ⁄ x²
• D) (ln(x) - 1) ⁄ x²

View Answer & Solution

Correct Answer: C

Solution: Let the top function be u = ln(x) → u' = 1/x and the bottom function be v = x → v' = 1.
Applying the Quotient Rule (vu' - uv') / v² gives:
[x(1/x) - ln(x)(1)] ⁄ x² = (1 - ln(x)) ⁄ x².

Question 8 (Chain Rule): Differentiate f(x) = (2x + 3)⁴.

• A) 4(2x + 3)³
• B) 8(2x + 3)³
• C) 8(2x + 3)⁴
• D) 2(2x + 3)³

View Answer & Solution

Correct Answer: B

Solution: Differentiate the outside power expression first, leaving the inside expression intact: 4(2x+3)³. Then multiply by the derivative of the inside expression, ^d⁄_dx(2x+3) = 2. This results in 4(2x+3)³ · 2 = 8(2x+3)³.

Question 9 (Exponential Rule): Find the derivative of f(x) = e^5x.

• A) e^5x
• B) 5e^x
• C) 5e^5x
• D) (1/5)e^5x

View Answer & Solution

Correct Answer: C

Solution: The derivative of e^u is e^u · u'. Here, u = 5x and its derivative is 5. Therefore, the overall derivative is e^5x · 5 = 5e^5x.

Question 10 (Logarithmic Rule): Find the derivative of f(x) = ln(4x).

• A) 1 ⁄ x
• B) 1 ⁄ 4x
• C) 4 ⁄ x
• D) 4ln(x)

View Answer & Solution

Correct Answer: A

Solution: Using the chain rule version of the log rule: ^d⁄_dx(ln(u)) = (1/u) · u'. Substituting u = 4x and u' = 4, we get (1/4x) · 4 = 4/4x = 1/x.

Solving Problems On Linear Graphs

 

Use the graph below to answer the following questions 

1. What is the y-intercept of the straight line shown in the graph?

• A) (1, 0)
• B) (0, 1)
• C) (0, -1)
• D) (-0.5, 0)

Correct Answer: B) (0, 1)

Solution: The y-intercept is the point where the line crosses the vertical y-axis (where x = 0). Looking closely at the graph, the line crosses the y-axis exactly at 1. The graph explicitly labels this coordinate as (0, 1).

2. What is the slope (gradient) of the line represented in the graph?

• A) 1
• B) -2
• C) 2
• D) 1/2

Correct Answer: C) 2

Solution: To find the slope (m), pick any two points marked with an "X" on the graph. For example, using (0, 1) and (1, 3):
m = (y₂ - y₁) / (x₂ - x₁)
m = (3 - 1) / (1 - 0) = 2 / 1 = 2.

Alternatively, observe that for every 1 unit you move to the right along the x-axis, the line rises by 2 units along the y-axis (Rise/Run = 2/1 = 2).

3. Which of the following equations perfectly defines the straight line in the graph?

• A) y = x + 1
• B) y = 2x + 1
• C) y = 2x - 1
• D) y = -2x + 1

Correct Answer: B) y = 2x + 1

Solution: The slope-intercept form of a linear equation is given by: y = mx + c, where m is the slope and c is the y-intercept. From the previous questions, the slope m = 2 and the y-intercept c = 1. Substituting these values into the formula yields: y = 2x + 1.

4. Based on the line's trajectory and markings in the graph, at what value of x does the line cross the horizontal x-axis?

• A) x = -0.5
• B) x = -1
• C) x = 0
• D) x = 0.5

Correct Answer: A) x = -0.5

Solution: The x-intercept occurs where y = 0. Using the equation of the line (y = 2x + 1), set y to 0 and solve for x:
0 = 2x + 1
-1 = 2x
x = -1/2 = -0.5.

Visually checking the graph confirms the blue line crosses the horizontal axis exactly halfway between 0 and -1.

5. Which of the following ordered pairs is NOT a point marked with an orange "X" on the line in the graph?

• A) (2, 5)
• B) (-1, -1)
• C) (-3, -5)
• D) (-2, -4)

Correct Answer: D) (-2, -4)

Solution: Check the options using the line's equation y = 2x + 1:
• For x = 2: y = 2(2) + 1 = 5 → (2, 5) is marked.
• For x = -1: y = 2(-1) + 1 = -1 → (-1, -1) is marked.
• For x = -3: y = 2(-3) + 1 = -5 → (-3, -5) is marked.
• For x = -2: y = 2(-2) + 1 = -3 → The marked point on the line is (-2, -3). Therefore, (-2, -4) does not lie on the line.