Math Problem Statement

can you make 5 examples of chain rule for me

Solution

Sure! Here are five examples that demonstrate the chain rule in calculus. The chain rule is used to differentiate composite functions, i.e., functions that are made by combining two or more simpler functions.


Example 1:

Find the derivative of f(x)=(3x2+4x)5f(x) = (3x^2 + 4x)^5.

Solution:
Let u=3x2+4xu = 3x^2 + 4x, so f(x)=u5f(x) = u^5.

Using the chain rule, we differentiate f(x)f(x) as follows:

f(x)=5u4dudxf'(x) = 5u^4 \cdot \frac{du}{dx}

Now differentiate u=3x2+4xu = 3x^2 + 4x:

dudx=6x+4\frac{du}{dx} = 6x + 4

Thus, the derivative of f(x)f(x) is:

f(x)=5(3x2+4x)4(6x+4)f'(x) = 5(3x^2 + 4x)^4 \cdot (6x + 4)


Example 2:

Find the derivative of g(x)=sin(2x3+3x)g(x) = \sin(2x^3 + 3x).

Solution:
Let u=2x3+3xu = 2x^3 + 3x, so g(x)=sin(u)g(x) = \sin(u).

Using the chain rule:

g(x)=cos(u)dudxg'(x) = \cos(u) \cdot \frac{du}{dx}

Now differentiate u=2x3+3xu = 2x^3 + 3x:

dudx=6x2+3\frac{du}{dx} = 6x^2 + 3

Thus, the derivative of g(x)g(x) is:

g(x)=cos(2x3+3x)(6x2+3)g'(x) = \cos(2x^3 + 3x) \cdot (6x^2 + 3)


Example 3:

Find the derivative of h(x)=e3x2+5xh(x) = e^{3x^2 + 5x}.

Solution:
Let u=3x2+5xu = 3x^2 + 5x, so h(x)=euh(x) = e^u.

Using the chain rule:

h(x)=eududxh'(x) = e^u \cdot \frac{du}{dx}

Now differentiate u=3x2+5xu = 3x^2 + 5x:

dudx=6x+5\frac{du}{dx} = 6x + 5

Thus, the derivative of h(x)h(x) is:

h(x)=e3x2+5x(6x+5)h'(x) = e^{3x^2 + 5x} \cdot (6x + 5)


Example 4:

Find the derivative of f(x)=ln(4x2+1)f(x) = \ln(4x^2 + 1).

Solution:
Let u=4x2+1u = 4x^2 + 1, so f(x)=ln(u)f(x) = \ln(u).

Using the chain rule:

f(x)=1ududxf'(x) = \frac{1}{u} \cdot \frac{du}{dx}

Now differentiate u=4x2+1u = 4x^2 + 1:

dudx=8x\frac{du}{dx} = 8x

Thus, the derivative of f(x)f(x) is:

f(x)=14x2+18xf'(x) = \frac{1}{4x^2 + 1} \cdot 8x

Which simplifies to:

f(x)=8x4x2+1f'(x) = \frac{8x}{4x^2 + 1}


Example 5:

Find the derivative of y(x)=5x4+3xy(x) = \sqrt{5x^4 + 3x}.

Solution:
Let u=5x4+3xu = 5x^4 + 3x, so y(x)=u=u1/2y(x) = \sqrt{u} = u^{1/2}.

Using the chain rule:

y(x)=12u1/2dudxy'(x) = \frac{1}{2} u^{-1/2} \cdot \frac{du}{dx}

Now differentiate u=5x4+3xu = 5x^4 + 3x:

dudx=20x3+3\frac{du}{dx} = 20x^3 + 3

Thus, the derivative of y(x)y(x) is:

y(x)=12(5x4+3x)1/2(20x3+3)y'(x) = \frac{1}{2} (5x^4 + 3x)^{-1/2} \cdot (20x^3 + 3)


Summary of Results:

  1. f(x)=5(3x2+4x)4(6x+4)f'(x) = 5(3x^2 + 4x)^4 \cdot (6x + 4)
  2. g(x)=cos(2x3+3x)(6x2+3)g'(x) = \cos(2x^3 + 3x) \cdot (6x^2 + 3)
  3. h(x)=e3x2+5x(6x+5)h'(x) = e^{3x^2 + 5x} \cdot (6x + 5)
  4. f(x)=8x4x2+1f'(x) = \frac{8x}{4x^2 + 1}
  5. y(x)=12(5x4+3x)1/2(20x3+3)y'(x) = \frac{1}{2} (5x^4 + 3x)^{-1/2} \cdot (20x^3 + 3)

Would you like a more detailed explanation of any specific example or concept? Let me know if you'd like further clarification.

Further Questions:

  1. How do you apply the chain rule when dealing with trigonometric functions?
  2. What is the difference between the product rule and the chain rule?
  3. Can the chain rule be applied to higher-order derivatives?
  4. What happens when you apply the chain rule to a nested function?
  5. How do you chain multiple functions together in a single expression?

Tip:

When using the chain rule, always start by identifying the outermost function and then work your way inward to find all the "u" substitutions.

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Math Problem Analysis

Mathematical Concepts

Calculus
Chain Rule
Differentiation
Composite Functions

Formulas

f'(x) = g'(u) * u'(x)
d/dx[e^u] = e^u * u'(x)
d/dx[sin(u)] = cos(u) * u'(x)
d/dx[ln(u)] = 1/u * u'(x)
d/dx[u^n] = n * u^(n-1) * u'(x)

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12