Understanding Derivatives
Before diving into specific problems, it's crucial to grasp the concept of a derivative. A derivative measures how a function changes as its input changes. Formally, if \( f(x) \) is a function, the derivative \( f'(x) \) at a point \( x \) can be defined as:
\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]
This definition encapsulates the idea of the instantaneous rate of change of the function at a specific point. Derivatives have numerous applications in physics, engineering, and economics, making them a vital part of calculus.
Basic Problems
Problem 1: Find the derivative of a polynomial function
Problem Statement: Find the derivative of the function \( f(x) = 3x^4 - 5x^3 + 2x - 7 \).
Solution:
To find the derivative of the polynomial, we apply the power rule, which states that if \( f(x) = ax^n \), then \( f'(x) = n \cdot ax^{n-1} \).
1. Differentiate each term:
- The derivative of \( 3x^4 \) is \( 12x^3 \).
- The derivative of \( -5x^3 \) is \( -15x^2 \).
- The derivative of \( 2x \) is \( 2 \).
- The derivative of \( -7 \) is \( 0 \).
2. Combine the derivatives:
\[
f'(x) = 12x^3 - 15x^2 + 2
\]
Problem 2: Find the derivative of a trigonometric function
Problem Statement: Find the derivative of the function \( g(x) = \sin(x) + \cos(2x) \).
Solution:
To differentiate the function, we use the derivatives of sine and cosine:
1. The derivative of \( \sin(x) \) is \( \cos(x) \).
2. For \( \cos(2x) \), we also apply the chain rule:
- The derivative of \( \cos(u) \) is \( -\sin(u) \), and \( u = 2x \) gives a factor of \( 2 \).
Putting it all together:
1. Differentiate \( \sin(x) \):
\[
\frac{d}{dx}[\sin(x)] = \cos(x)
\]
2. Differentiate \( \cos(2x) \):
\[
\frac{d}{dx}[\cos(2x)] = -\sin(2x) \cdot 2 = -2\sin(2x)
\]
3. Combine the results:
\[
g'(x) = \cos(x) - 2\sin(2x)
\]
Intermediate Problems
Problem 3: Find the derivative using the product rule
Problem Statement: Find the derivative of the function \( h(x) = (2x^3)(\ln(x)) \).
Solution:
To differentiate a product of two functions, we use the product rule:
\[
(fg)' = f'g + fg'
\]
Let:
- \( f(x) = 2x^3 \)
- \( g(x) = \ln(x) \)
1. Differentiate \( f(x) \):
\[
f'(x) = 6x^2
\]
2. Differentiate \( g(x) \):
\[
g'(x) = \frac{1}{x}
\]
3. Apply the product rule:
\[
h'(x) = f'(x)g(x) + f(x)g'(x) = (6x^2)(\ln(x)) + (2x^3)\left(\frac{1}{x}\right)
\]
4. Simplify:
\[
h'(x) = 6x^2 \ln(x) + 2x^2 = 2x^2(3\ln(x) + 1)
\]
Problem 4: Find the derivative using the quotient rule
Problem Statement: Find the derivative of the function \( k(x) = \frac{x^2 + 1}{x - 1} \).
Solution:
To differentiate a quotient of two functions, we use the quotient rule:
\[
\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}
\]
Let:
- \( f(x) = x^2 + 1 \)
- \( g(x) = x - 1 \)
1. Differentiate \( f(x) \):
\[
f'(x) = 2x
\]
2. Differentiate \( g(x) \):
\[
g'(x) = 1
\]
3. Apply the quotient rule:
\[
k'(x) = \frac{(2x)(x - 1) - (x^2 + 1)(1)}{(x - 1)^2}
\]
4. Simplify:
\[
k'(x) = \frac{2x^2 - 2x - x^2 - 1}{(x - 1)^2} = \frac{x^2 - 2x - 1}{(x - 1)^2}
\]
Advanced Problems
Problem 5: Implicit differentiation
Problem Statement: Use implicit differentiation to find \( \frac{dy}{dx} \) for the equation \( x^2 + y^2 = 25 \).
Solution:
1. Differentiate both sides with respect to \( x \):
\[
\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)
\]
This gives:
\[
2x + 2y\frac{dy}{dx} = 0
\]
2. Solve for \( \frac{dy}{dx} \):
\[
2y\frac{dy}{dx} = -2x \Rightarrow \frac{dy}{dx} = -\frac{x}{y}
\]
Problem 6: Higher order derivatives
Problem Statement: Find the second derivative of the function \( f(x) = x^3 - 3x^2 + 2x \).
Solution:
1. First, find the first derivative:
\[
f'(x) = 3x^2 - 6x + 2
\]
2. Next, find the second derivative:
\[
f''(x) = 6x - 6
\]
3. This completes the solution, giving us \( f''(x) = 6(x - 1) \).
Conclusion
Differential calculus serves as a foundational tool in mathematics and its applications. By solving problems ranging from basic derivatives of polynomials to more complex scenarios involving implicit differentiation and higher-order derivatives, students can build a robust understanding of how to analyze and interpret changes in functions. The problems presented in this article illustrate various techniques and rules, such as the power rule, product rule, quotient rule, and implicit differentiation. Mastery of these concepts is essential for tackling more advanced topics in calculus and related fields.
Frequently Asked Questions
What is the derivative of the function f(x) = 3x^2 + 5x - 7?
The derivative f'(x) = 6x + 5.
How do you find the critical points of the function f(x) = x^3 - 3x^2 + 4?
To find critical points, first find the derivative f'(x) = 3x^2 - 6x. Set the derivative to zero: 3x^2 - 6x = 0, which gives x(x - 2) = 0. Thus, critical points are x = 0 and x = 2.
What is the second derivative of f(x) = 2x^3 - 4x?
The first derivative is f'(x) = 6x^2 - 4. The second derivative is f''(x) = 12x.
How can you apply the product rule to differentiate f(x) = (2x)(x^2 + 1)?
Using the product rule, f'(x) = (2x)'(x^2 + 1) + (2x)(x^2 + 1)'. This gives f'(x) = 2(x^2 + 1) + 2x(2x) = 2x^2 + 2 + 4x^2 = 6x^2 + 2.
What is the chain rule and how do you use it for f(x) = sin(3x^2)?
The chain rule states that if you have a composite function, f(g(x)), then f'(x) = f'(g(x)) g'(x). For f(x) = sin(3x^2), let g(x) = 3x^2. So f'(x) = cos(3x^2) g'(x) = cos(3x^2) 6x.
How do you find the slope of the tangent line to the curve y = x^2 at the point (2, 4)?
First, find the derivative y' = 2x. Then, evaluate it at x = 2: y'(2) = 2(2) = 4. Therefore, the slope of the tangent line at (2, 4) is 4.
What is the derivative of the function f(x) = e^(2x)?
Using the chain rule, the derivative is f'(x) = 2e^(2x).
How do you determine if a function has a local minimum or maximum using the first derivative test?
Identify critical points where f'(x) = 0 or is undefined. Analyze the sign of f'(x) before and after each critical point. If f'(x) changes from positive to negative, there is a local maximum; if it changes from negative to positive, there is a local minimum.
What is the derivative of f(x) = ln(x^2 + 1)?
Using the chain rule, f'(x) = (1/(x^2 + 1)) (2x) = 2x/(x^2 + 1).
