Understanding Differential Calculus
Differential calculus focuses on the concept of the derivative, which measures how a function changes as its input changes. The derivative of a function at a point gives the slope of the tangent line to the curve of the function at that point.
Key Concepts
1. Derivative Definition: The derivative of a function f(x) at a point x = a is defined as:
\[
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}
\]
2. Notation:
- \( f'(x) \) for the derivative.
- \( \frac{dy}{dx} \) for the derivative of y with respect to x.
- \( Df \) or \( \frac{d}{dx} f(x) \).
3. Basic Derivative Rules:
- Power Rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
- Product Rule: If \( f(x) = u(x)v(x) \), then \( f'(x) = u'v + uv' \).
- Quotient Rule: If \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'v - uv'}{v^2} \).
- Chain Rule: If \( f(x) = g(h(x)) \), then \( f'(x) = g'(h(x)) \cdot h'(x) \).
Common Differential Calculus Problems
Let’s explore some typical problems encountered in differential calculus along with their solutions.
Problem 1: Finding the Derivative
Problem Statement: Find the derivative of the function \( f(x) = 3x^4 - 2x^3 + x - 7 \).
Solution:
To find the derivative, apply the power rule:
1. Differentiate each term:
- The derivative of \( 3x^4 \) is \( 12x^3 \).
- The derivative of \( -2x^3 \) is \( -6x^2 \).
- The derivative of \( x \) is \( 1 \).
- The derivative of \( -7 \) is \( 0 \).
2. Combine the results:
\[
f'(x) = 12x^3 - 6x^2 + 1
\]
Problem 2: Using the Product Rule
Problem Statement: Use the product rule to differentiate \( f(x) = (2x^3)(\sin x) \).
Solution:
1. Identify \( u = 2x^3 \) and \( v = \sin x \).
2. Find \( u' = 6x^2 \) and \( v' = \cos x \).
3. Apply the product rule:
\[
f'(x) = u'v + uv' = (6x^2)(\sin x) + (2x^3)(\cos x)
\]
4. Thus,
\[
f'(x) = 6x^2 \sin x + 2x^3 \cos x
\]
Problem 3: Finding Critical Points
Problem Statement: Determine the critical points of the function \( f(x) = x^3 - 3x^2 + 4 \).
Solution:
1. First, find the derivative:
\[
f'(x) = 3x^2 - 6x
\]
2. Set the derivative equal to zero:
\[
3x^2 - 6x = 0 \implies 3x(x - 2) = 0
\]
3. Solve for x:
- \( x = 0 \)
- \( x = 2 \)
4. Therefore, the critical points are \( x = 0 \) and \( x = 2 \).
Problem 4: Finding Local Extrema
Problem Statement: Using the critical points found in Problem 3, determine the nature of these points (local minima or maxima).
Solution:
1. Use the second derivative test:
\[
f''(x) = 6x - 6
\]
2. Evaluate \( f''(x) \) at the critical points:
- For \( x = 0 \):
\[
f''(0) = 6(0) - 6 = -6 \quad (\text{local maximum})
\]
- For \( x = 2 \):
\[
f''(2) = 6(2) - 6 = 6 \quad (\text{local minimum})
\]
3. Conclusion: \( x = 0 \) is a local maximum, and \( x = 2 \) is a local minimum.
Problem 5: Application of Derivatives - Optimization
Problem Statement: A rectangular garden is to be made with a perimeter of 40 meters. What dimensions will maximize the area?
Solution:
1. Let the length be \( l \) and the width be \( w \). The perimeter constraint gives:
\[
2l + 2w = 40 \implies l + w = 20 \implies w = 20 - l
\]
2. The area \( A \) is given by:
\[
A = lw = l(20 - l) = 20l - l^2
\]
3. Find the derivative:
\[
A' = 20 - 2l
\]
4. Set the derivative equal to zero to find critical points:
\[
20 - 2l = 0 \implies l = 10
\]
5. Therefore, \( w = 20 - 10 = 10 \).
6. The dimensions that maximize the area are \( 10 \) meters by \( 10 \) meters.
Conclusion
In summary, differential calculus problems and solutions encompass a wide range of applications, from finding derivatives to optimizing functions. By mastering the techniques of differentiation, one can tackle various mathematical challenges effectively. The problems discussed here illustrate some of the fundamental concepts of differential calculus, providing a solid foundation for further study in this essential area of mathematics. As you continue to explore calculus, remember that practice is key to gaining proficiency in solving these types of problems.
Frequently Asked Questions
What is the derivative of the function f(x) = 3x^2 + 5x - 7?
The derivative f'(x) is 6x + 5.
How do you find the critical points of the function f(x) = x^3 - 3x^2 + 4?
To find the critical points, first compute the derivative f'(x) = 3x^2 - 6x. Set it to zero: 3x^2 - 6x = 0, which gives x(x - 2) = 0. Thus, the critical points are x = 0 and x = 2.
What is the second derivative test for identifying local extrema?
The second derivative test states that if f''(x) > 0 at a critical point, then f has a local minimum there; if f''(x) < 0, then f has a local maximum; if f''(x) = 0, the test is inconclusive.
How do you apply the chain rule to differentiate the function g(x) = sin(2x^2)?
Using the chain rule, g'(x) = cos(2x^2) (4x) = 4x cos(2x^2).
What is the product rule and how is it applied to the function h(x) = (x^2 + 1)(x - 3)?
The product rule states that if h(x) = u(x)v(x), then h'(x) = u'(x)v(x) + u(x)v'(x). For h(x), u(x) = x^2 + 1 and v(x) = x - 3. Thus, h'(x) = (2x)(x - 3) + (x^2 + 1)(1) = 2x^2 - 6x + x^2 + 1 = 3x^2 - 6x + 1.
