Tuesday, 9 June 2026

Mathematical Induction Made Easy

Understand Mathematical Induction

Line-by-Line Breakdown of the Inductive Proof

Let's strip away the intimidating symbols and break down the algebra from Step 3 in simple language. Think of this process like a puzzle where we are swapping out large, messy pieces for smaller, cleaner ones.

Line 1: 1 + 3 + 5 + ... + (2k - 1) + [2(k+1) - 1]

What it means: This is simply a long list of consecutive odd numbers written out horizontally.
The front half, 1 + 3 + 5 + ... + (2k - 1), is the collection of all the odd numbers up to a random stopping point in line called k.
The part wrapped in square brackets, [2(k+1) - 1], is just the very next odd number standing right behind it.

Line 2: = k² + [2(k+1) - 1]

What it means: This is where the magic substitution happens. Remember our assumption in Step 2? We decided that adding up everything up to position k is equal to k².
Because we assume that rule holds true, we sweep that whole bulky front train of numbers off the table and drop down a clean k² in its place.
The extra odd number at the tail end stays right where it is, waiting for its turn to be simplified.

Line 3: = k² + [2k + 2 - 1]

What it means: Now we focus completely on breaking down the messy stuff trapped inside the square brackets.
We look at 2(k+1) and distribute the multiplier: 2 times k gives us 2k, and 2 times 1 gives us 2.
The minus 1 simply sits quietly at the end of the line.

Line 4: = k² + 2k + 1

What it means: This is simple elementary cleanup. Inside the brackets, we combine the normal numbers: 2 minus 1 equals 1.
We discard the square brackets entirely because they aren't needed anymore, leaving us with a beautiful, clean quadratic expression: k² + 2k + 1.

Line 5: = (k + 1)²

What it means: This is our grand finale match. If you recall basic algebra factorization, the expression k² + 2k + 1 splits perfectly into two identical pieces: (k + 1)(k + 1). We write this cleanly as (k + 1)².
Why this is a total win: Our initial goal was to see if adding up (k+1) odd numbers would give us that number of items squared. By letting the logic flow naturally, we proved it perfectly!

Simple Example On Mathematical Induction

Think of Mathematical Induction as the ultimate way to knock over an infinite line of dominoes.

Instead of walking down the line and pushing every single domino over one by one (which would take forever), you only need to prove two things:

The First Domino Falls: You can actually push the very first domino over.
The Chain Reaction Works: If any given domino falls, it is guaranteed to knock over the next one.

If you prove both, you've proven that the entire infinite line of dominoes will fall.

The Example: The Odd Number Ladder

Let's say someone tells you a trick: "If you add up consecutive odd numbers starting from 1, the answer is always a perfect square."

Let's test it out to see if they are right:

Add the first 1 odd number: 1 = 1²
Add the first 2 odd numbers: 1 + 3 = 4 = 2²
Add the first 3 odd numbers: 1 + 3 + 5 = 9 = 3²
Add the first 4 odd numbers: 1 + 3 + 5 + 7 = 16 = 4²

It seems to work perfectly! The rule or formula for this pattern is:

1 + 3 + 5 + ... + (2n - 1) = n²

But how do we prove this works for the first 100 odd numbers? Or the first million? Or forever? This is where mathematical induction saves the day in three simple steps.

Step 1: The Base Case (The First Domino)

First, we must prove the formula works for the very first number, where n = 1.

Left side (just the first number): 1
Right side (using the formula n² where n=1): 1² = 1

Both sides match! The first domino falls.

Step 2: The Inductive Hypothesis (The "What If?")

Now, we assume the trick works for some random step in the middle of the line. Let's call this step k.

We just assume this mathematical statement is completely true:

1 + 3 + 5 + ... + (2k - 1) = k²

This is our setup. We are looking at the line of dominoes and saying, "Let's imagine domino k just fell."

Step 3: The Inductive Step (The Chain Reaction)

Now we have to prove that because domino k fell, the next domino, domino k + 1, must also fall.

If we write out the sum for k + 1 terms, it looks like this:

1 + 3 + 5 + ... + (2k - 1) + [2(k+1) - 1]

Look closely at that long string of numbers. Notice that the front part of it is exactly our Step 2 setup! Because we assumed that front part equals k², we can swap it out:

= k² + [2(k+1) - 1]

Now, let's simplify the stuff inside the brackets:

= k² + [2k + 2 - 1]

= k² + 2k + 1

If you remember basic high school algebra, k² + 2k + 1 can be factored perfectly into a square:

= (k + 1)²

The Grand Conclusion

Look at what we just accomplished! By assuming the formula worked for k, we did a little algebra and showed that the next step automatically simplified to (k + 1)².

Because we proved it works for 1 (Step 1), it automatically triggers it to work for 2.
Because it works for 2, it automatically triggers it to work for 3.
Because it works for 3, it triggers 4... and so on, all the way to infinity.

You've just knocked down every single domino in existence using math induction!

Monday, 8 June 2026

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:

The first domino falls.

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.

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

Sunday, 7 June 2026

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.

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).

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:

Differentiate both sides of the equation with respect to x.
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.
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

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