Integral Calculus - Leibniz rule for differentiation and curve passing through a point

• In this lecture, we will discuss the Leibniz rule for differentiation and how to find a curve passing through a given point using this rule.

• Leibniz rule, also known as the generalized product rule, is a powerful tool in integral calculus.

• It allows us to find the derivative of a product of two functions.

• The rule states that if we have two functions, say f(x) and g(x), then the derivative of their product f(x)g(x) is given by: f’(x)g(x) + f(x)g’(x)

• This rule can be extended to more than two functions by applying the product rule iteratively.

• Let’s understand this rule with an example: If f(x) = sin(x) and g(x) = x^2, find the derivative of their product f(x)g(x).

• Solution: Applying the Leibniz rule, we have: f’(x)g(x) + f(x)g’(x) = cos(x)x^2 + sin(x)(2x)

• Simplifying further, we get: cos(x)x^2 + 2xsin(x)

• Therefore, the derivative of the product f(x)g(x) is cos(x)x^2 + 2xsin(x).

Curve passing through a given point

• The Leibniz rule can also be used to find a curve passing through a given point.

• Let’s consider the equation of a curve passing through the point (1,2) in the form y = f(x).

• Using the Leibniz rule, we can find the equation of the curve by finding the antiderivative of the given function.

• Let’s take an example to illustrate this concept: Find the equation of a curve passing through the point (1,2) if its derivative is given by f’(x) = 2x.

• Solution: We need to find the antiderivative of f’(x) = 2x. Denoting the antiderivative as F(x), we have: F(x) = ∫2x dx = x^2 + C Applying the point (1,2), we can substitute the values of x and y in the equation: 2 = 1^2 + C

• Solving for C, we get C = 1.

• Therefore, the equation of the curve passing through the point (1,2) is y = x^2 + 1.

Recap

• The Leibniz rule, also known as the generalized product rule, is used to find the derivative of a product of two or more functions.
• The rule states that the derivative of the product f(x)g(x) is given by f’(x)g(x) + f(x)g’(x).
• The Leibniz rule can also be used to find the equation of a curve passing through a given point.
• To find the equation of the curve, we need to integrate the given derivative and solve for the constant of integration using the given point.
• Let’s practice a few more examples to solidify our understanding of the Leibniz rule and finding curves passing through given points.
• It is important to understand and apply the Leibniz rule correctly, as it plays a significant role in various applications of calculus.

Example 1

• Find the derivative of the product of the functions f(x) = x^3 - 2x and g(x) = sin(x).
• Solution: Applying the Leibniz rule, we have: f’(x)g(x) + f(x)g’(x) = (3x^2 - 2)sin(x) + (x^3 - 2x)cos(x) Simplifying further, we get: (3x^2 - 2)sin(x) + (x^3 - 2x)cos(x)
• Therefore, the derivative of the product f(x)g(x) is (3x^2 - 2)sin(x) + (x^3 - 2x)cos(x).

Example 2

• Find the equation of the curve passing through the point (2,5) if its derivative is given by f’(x) = 3x^2 - 4x.
• Solution: We need to find the antiderivative of f’(x) = 3x^2 - 4x. Denoting the antiderivative as F(x), we have: F(x) = ∫(3x^2 - 4x) dx = x^3 - 2x^2 + C Applying the point (2,5), we can substitute the values of x and y in the equation: 5 = (2)^3 - 2(2)^2 + C
• Solving for C, we get C = 5 - 8 + 8 = 5.
• Therefore, the equation of the curve passing through the point (2,5) is y = x^3 - 2x^2 + 5.

Additional Practice

• Let’s practice a few more problems to reinforce our understanding of the Leibniz rule and finding curves passing through given points.
• Problem 1: Find the derivative of the product of the functions f(x) = 2x^2 - 3 and g(x) = e^x.
• Problem 2: Find the equation of the curve passing through the point (3,4) if its derivative is given by f’(x) = cos(x).

11. Examples of the Leibniz rule:

• Find the derivative of the product of the functions f(x) = x^2 - 3x and g(x) = cos(x).
• Find the derivative of the product of the functions f(x) = e^x and g(x) = ln(x).
• Find the derivative of the product of the functions f(x) = sqrt(x) and g(x) = 1/x.
• Find the derivative of the product of the functions f(x) = x^3 and g(x) = sin(x).
• Find the derivative of the product of the functions f(x) = 2x and g(x) = e^(2x).

12. Solving a curve passing through a given point:

• Let’s consider the equation of a curve passing through the point (0,1).
• Given the derivative f’(x) = 3x^2 + 2x, we need to find the equation of the curve.
• Integrating the derivative, we get F(x) = x^3 + x^2 + C.
• Substituting the values of x and y using the given point, we get 1 = (0)^3 + (0)^2 + C.
• Solving for C, we find C = 1.
• Therefore, the equation of the curve passing through the point (0,1) is y = x^3 + x^2 + 1.

13. Solving a curve passing through a given point (continued):

• Let’s consider the equation of a curve passing through the point (-1,4).
• Given the derivative f’(x) = 2x + 3, we need to find the equation of the curve.
• Integrating the derivative, we get F(x) = x^2 + 3x + C.
• Substituting the values of x and y using the given point, we get 4 = (-1)^2 + 3(-1) + C.
• Solving for C, we find C = 6.
• Therefore, the equation of the curve passing through the point (-1,4) is y = x^2 + 3x + 6.

14. Summary of the Leibniz rule:

• The Leibniz rule allows us to find the derivative of a product of functions.
• The rule states that the derivative of the product f(x)g(x) is f’(x)g(x) + f(x)g’(x).
• This rule can be applied to find the derivatives of complicated functions.
• The Leibniz rule can also be used to find the equation of a curve passing through a given point.
• To find the equation of the curve, we need the derivative of the function and the given point.

15. Tips for applying the Leibniz rule:

• Make sure to correctly identify the functions f(x) and g(x) in the product.
• Differentiate each function individually and find their derivatives.
• Substitute the derivatives and the original functions in the Leibniz rule formula.
• Simplify the expression to find the derivative of the product.
• When finding a curve passing through a given point, integrate the derivative and solve for the constant of integration using the given point.

16. Key takeaways:

• The Leibniz rule is a powerful tool in integral calculus for finding derivatives of products.
• It can be extended to find the derivatives of more than two functions.
• The Leibniz rule can also be used to find the equation of a curve passing through a given point.
• To find the equation, integrate the given derivative and solve for the constant of integration using the given point.

17. Practice problems:

• Find the derivative of the product of the functions f(x) = x^4 - 3x^2 + 2 and g(x) = e^x.
• Find the equation of the curve passing through the point (-2,3) if its derivative is given by f’(x) = 4x^3 + x.
• Find the derivative of the product of the functions f(x) = ln(x) and g(x) = 1/x^2.
• Find the equation of the curve passing through the point (4,5) if its derivative is given by f’(x) = 2cos(x).
• Find the derivative of the product of the functions f(x) = 3x - 1 and g(x) = 2x^2 - 4x + 5.

21. Example 3:

• Find the derivative of the product of the functions f(x) = 2x^3 - 5x and g(x) = ln(x).
• Solution:
  • Applying the Leibniz rule, we have: f’(x)g(x) + f(x)g’(x) = (6x^2 - 5)ln(x) + (2x^3 - 5x)(1/x)
  • Simplifying further, we get: (6x^2 - 5)ln(x) + 2x^2 - 5
• Therefore, the derivative of the product f(x)g(x) is (6x^2 - 5)ln(x) + 2x^2 - 5.

22. Example 4:

• Find the equation of the curve passing through the point (-2,3) if its derivative is given by f’(x) = 4x^3 + x.
• Solution:
  • We need to find the antiderivative of f’(x) = 4x^3 + x. Denoting the antiderivative as F(x), we have: F(x) = ∫(4x^3 + x) dx = x^4/4 + x^2/2 + C
  • Applying the point (-2,3), we can substitute the values of x and y in the equation: 3 = (-2)^4/4 + (-2)^2/2 + C
  • Solving for C, we get C = 3 - 4 + 2 = 1.
• Therefore, the equation of the curve passing through the point (-2,3) is y = x^4/4 + x^2/2 + 1.

23. Example 5:

• Find the derivative of the product of the functions f(x) = sqrt(x) and g(x) = 1/x.
• Solution:
  • Applying the Leibniz rule, we have: f’(x)g(x) + f(x)g’(x) = (1/2sqrt(x))(1/x) + sqrt(x)(-1/x^2)
  • Simplifying further, we get: 1/2xsqrt(x) - sqrt(x)/x^2
• Therefore, the derivative of the product f(x)g(x) is 1/2xsqrt(x) - sqrt(x)/x^2.

24. Example 6:

• Find the equation of the curve passing through the point (4,5) if its derivative is given by f’(x) = 2cos(x).
• Solution:
  • We need to find the antiderivative of f’(x) = 2cos(x). Denoting the antiderivative as F(x), we have: F(x) = ∫2cos(x) dx = 2sin(x) + C
  • Applying the point (4,5), we can substitute the values of x and y in the equation: 5 = 2sin(4) + C
  • Solving for C, we get C = 5 - 2sin(4).
• Therefore, the equation of the curve passing through the point (4,5) is y = 2sin(x) + (5 - 2sin(4)).

25. Example 7:

• Find the derivative of the product of the functions f(x) = 3x - 1 and g(x) = 2x^2 - 4x + 5.
• Solution:
  • Applying the Leibniz rule, we have: f’(x)g(x) + f(x)g’(x) = (3)(2x^2 - 4x + 5) + (3x - 1)(4x - 4)
  • Simplifying further, we get: 6x^2 - 12x + 15 + 12x^2 - 12x - 4
  • Combining like terms, we get: 18x^2 - 24
• Therefore, the derivative of the product f(x)g(x) is 18x^2 - 24.

26. Summary of the Leibniz rule (continued):

• The Leibniz rule is a powerful tool in integral calculus for finding derivatives of products.
• It can be extended to find the derivatives of more than two functions.
• The Leibniz rule can also be used to find the equation of a curve passing through a given point.
• To find the equation, integrate the given derivative and solve for the constant of integration using the given point.

27. Tips for applying the Leibniz rule (continued):

• Make sure to correctly identify the functions f(x) and g(x) in the product.
• Differentiate each function individually and find their derivatives.
• Substitute the derivatives and the original functions in the Leibniz rule formula.
• Simplify the expression to find the derivative of the product.
• When finding a curve passing through a given point, integrate the derivative and solve for the constant of integration using the given point.

28. Summary of key concepts:

• The Leibniz rule is used to find the derivative of the product of two or more functions.
• The rule states that the derivative of the product f(x)g(x) is given by f’(x)g(x) + f(x)g’(x).
• The Leibniz rule can also be used to find the equation of a curve passing through a given point.
• To find the equation, integrate the given derivative and solve for the constant of integration using the given point.

29. Practice problems:

• Find the derivative of the product of the functions f(x) = x^2 - 5x and g(x) = e^x.
• Find the equation of the curve passing through the point (-3,2) if its derivative is given by f’(x) = 2x^2 + 3x.
• Find the derivative of the product of the functions f(x) = ln(x) and g(x) = 1/x^3.
• Find the equation of the curve passing through the point (1,4) if its derivative is given by f’(x) = 3sin(x).
• Find the derivative of the product of the functions f(x) = 3x + 1 and g(x) = e^(3x).

30. Summary and conclusion:

• The Leibniz rule is a powerful tool in calculus for finding derivatives of products.
• It can be used to find the derivative of a product of two or more functions.
• The rule states that the derivative of the product f(x)g(x) is given by f’(x)g(x) + f(x)g’(x).
• The Leibniz rule can also be used to find the equation of a curve passing through a given point.
• To find the equation, integrate the given derivative and solve for the constant of integration using the given point.
• Practice problems and examples are essential to reinforce understanding and apply the Leibniz rule effectively.
• Remember to simplify the expressions and solve for constants accurately.
• Apply the Leibniz rule correctly to solve more complex problems in calculus.