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The derivative of a function f(x) is denoted as f'(x) or df/dx
Geometrically, the derivative represents the slope of the tangent to a curve at a point
If y = f(x), then dy/dx = f'(x)
Derivative of xⁿ is nxⁿ⁻¹
Derivative of s∈ x is cos x
Derivative of cos x is -s∈ x
Derivative of tan x is sec²x
Derivative of eˣ is eˣ
Derivative of log(x) is 1/x
For a function y = f(g(x)), the derivative is given by dy/dx = f'(g(x)) * g'(x)
For a function y = f(x)/g(x), the derivative is given by dy/dx = (f'(x)g(x) - f(x)g'(x)) / g(x)²
The second derivative of a function f(x) is denoted as f''(x) or d²f/dx²
If f(x) is a function, then its derivative f'(x) represents the rate of change of the function with respect to x
Increasing function: f'(x) > 0
Decreasing function: f'(x) < 0
Maxima or Minima: f'(x) = 0
Inflection point: f''(x) = 0

🪤 The 5 Mistakes That Cost Marks

Not checking the doma∈ of the function before differentiating
Forgetting to apply the cha∈ rule ∈ composite functions
Not using the correct formula for the derivative of trigonometric functions
Not checking for the existence of the derivative at a point
Not using the second derivative test to determine the nature of the critical points

✏️ 3 Solved PYQs

Question 1: Find the derivative of the function f(x) = x³ s∈ x
Solution: Using the product rule, f'(x) = d(x³ s∈ x)/dx = x³ cos x + s∈ x * 3x² = x³ cos x + 3x² s∈ x
Question 2: Find the equation of the tangent to the curve y = x² + 3x - 2 at the point (1, 2)
Solution: First, find the derivative of the function y = x² + 3x - 2, which is dy/dx = 2x + 3
At the point (1, 2), the slope of the tangent is 2*1 + 3 = 5
The equation of the tangent is y - 2 = 5(x - 1)
Question 3: Find the maximum value of the function f(x) = x³ - 6x² + 9x + 2
Solution: First, find the critical points by taking the derivative and equating it to zero
f'(x) = 3x² - 12x + 9 = 0
Solving for x, we get x = 1 and x = 3
Now, use the second derivative test to determine the nature of the critical points
f''(x) = 6x - 12
At x = 1, f''(1) = 6 - 12 = -6 < 0, so x = 1 is a local maxima
At x = 3, f''(3) = 18 - 12 = 6 > 0, so x = 3 is a local minimum
The maximum value of the function is at x = 1, which is f(1) = 1 - 6 + 9 + 2 = 6

🧠 The One Thing Most Students Get Wrong

Most students get wrong the application of the cha∈ rule ∈ composite functions
They often forget to multiply the derivative of the outer function by the derivative of the inner function
For example, if y = s∈(x²), then dy/dx = cos(x²) * d(x²)/dx = cos(x²) * 2x
Another common mistake is not using the correct formula for the derivative of trigonometric functions
For example, the derivative of tan x is sec²x, not tan x

👁️ Ayush's Note

To solve application of derivatives problems, first identify the type of problem
If it is a maxima/minima problem, use the first derivative test or the second derivative test
If it is a tangent equation problem, use the point-slope form of a line
If it is a rate of change problem, use the concept of related rates
Always check the doma∈ of the function before differentiating
Always use the correct formula for the derivative of trigonometric functions
Practice, practice, practice, as application of derivatives is a topic that requires a lot of practice to master

🔁 Last 5 Minutes Box

Revision of important formulas and concepts
Practice of solving problems quickly and efficiently
Focus on the most common types of problems
Use of shortcuts and tricks to solve problems quickly
Stay calm and focused, and try to solve as many problems as possible ∈ the last 5 minutes

📝 Practice MCQs

1. Question: Find the derivative of the function f(x) = x² s∈ x

A) x² cos x + 2x s∈ x

B) x² cos x - 2x s∈ x

C) x² cos x + x s∈ x

D) x² cos x - x s∈ x

Answer: A) x² cos x + 2x s∈ x

2. Question: Find the equation of the tangent to the curve y = x³ - 2x² + x - 1 at the point (1, -1)

A) y + 1 = 2(x - 1)

B) y + 1 = 0(x - 1)

C) y + 1 = -2(x - 1)

D) y + 1 = 1(x - 1)

Answer: C) y + 1 = 0(x - 1)

3. Question: Find the maximum value of the function f(x) = x³ - 3x² + 2x + 1

A) 4

B) 5

C) 6

D) 7

Answer: B) 5

4. Question: Find the derivative of the function f(x) = eˣ s∈ x

A) eˣ s∈ x + eˣ cos x

B) eˣ s∈ x - eˣ cos x

C) eˣ s∈ x + eˣ tan x

D) eˣ s∈ x - eˣ tan x

Answer: A) eˣ s∈ x + eˣ cos x

5. Question: Find the equation of the tangent to the curve y = x² + 2x - 3 at the point (1, 0)

A) y - 0 = 4(x - 1)

B) y - 0 = 2(x - 1)

C) y - 0 = -4(x - 1)

D) y - 0 = -2(x - 1)

Answer: B) y - 0 = 4(x - 1)

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This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.

📚 Related Topics

Continue your revision with these related guides:

The derivative of a function f(x) is denoted as f'(x) or df/dx
Geometrically, the derivative represents the slope of the tangent to a curve at a point
If y = f(x), then dy/dx = f'(x)
Derivative of xⁿ is nxⁿ⁻¹
Derivative of s∈ x is cos x
Derivative of cos x is -s∈ x
Derivative of tan x is sec²x
Derivative of eˣ is eˣ
Derivative of log(x) is 1/x
For a function y = f(g(x)), the derivative is given by dy/dx = f'(g(x)) * g'(x)
For a function y = f(x)/g(x), the derivative is given by dy/dx = (f'(x)g(x) - f(x)g'(x)) / g(x)²
The second derivative of a function f(x) is denoted as f''(x) or d²f/dx²
If f(x) is a function, then its derivative f'(x) represents the rate of change of the function with respect to x
Increasing function: f'(x) > 0
Decreasing function: f'(x) < 0
Maxima or Minima: f'(x) = 0
Inflection point: f''(x) = 0