Math Problem Statement

Find the derivative of each function using the chain rule.

Solution

This is a Chain Rule worksheet, where you need to find the derivative of each function. I can guide you through the solution of these problems step by step. Let's go through the first few problems:

1. f(x)=(2x25x)3f(x) = (2x^2 - 5x)^3

We will use the chain rule for this function. Let u=2x25xu = 2x^2 - 5x, so f(x)=u3f(x) = u^3.

  • First, differentiate u3u^3 with respect to uu: ddu(u3)=3u2\frac{d}{du}(u^3) = 3u^2.
  • Then, differentiate u=2x25xu = 2x^2 - 5x with respect to xx: ddx(2x25x)=4x5\frac{d}{dx}(2x^2 - 5x) = 4x - 5.

Now, apply the chain rule: f(x)=3(2x25x)2(4x5)f'(x) = 3(2x^2 - 5x)^2 \cdot (4x - 5)

2. f(x)=5x32xf(x) = \sqrt{5x^3 - 2x}

We rewrite the square root as a power of 12\frac{1}{2}: f(x)=(5x32x)1/2f(x) = (5x^3 - 2x)^{1/2}.

  • Let u=5x32xu = 5x^3 - 2x, then f(x)=u1/2f(x) = u^{1/2}.
  • The derivative of u1/2u^{1/2} with respect to uu is 12u1/2\frac{1}{2}u^{-1/2}.
  • Differentiate u=5x32xu = 5x^3 - 2x with respect to xx: ddx(5x32x)=15x22\frac{d}{dx}(5x^3 - 2x) = 15x^2 - 2.

Now apply the chain rule: f(x)=12(5x32x)1/2(15x22)f'(x) = \frac{1}{2}(5x^3 - 2x)^{-1/2} \cdot (15x^2 - 2)

3. y=3sin(x3)y = 3\sin(x - 3)

For this one, use the chain rule combined with the derivative of sin(x)\sin(x).

  • Differentiate 3sin(x3)3\sin(x - 3) with respect to (x3)(x - 3): The derivative of sin(x)\sin(x) is cos(x)\cos(x), so we get 3cos(x3)3\cos(x - 3).

Thus, y=3cos(x3)y' = 3\cos(x - 3).

4. y=2cos(x2+2)y = -2\cos(x^2 + 2)

  • The derivative of cos(x)\cos(x) is sin(x)-\sin(x), so we differentiate 2cos(x2+2)-2\cos(x^2 + 2) with respect to (x2+2)(x^2 + 2).
  • First, differentiate cos(x2+2)\cos(x^2 + 2), which gives sin(x2+2)-\sin(x^2 + 2), then apply the chain rule:
  • Derivative of x2+2x^2 + 2 with respect to xx is 2x2x.

Now apply the chain rule: y=2(sin(x2+2))(2x)=4xsin(x2+2)y' = -2 \cdot (-\sin(x^2 + 2)) \cdot (2x) = 4x \sin(x^2 + 2)

5. g(x)=sin2(3x2)g(x) = \sin^2(3x^2)

This is the chain rule applied twice.

  • Let u=sin(3x2)u = \sin(3x^2), so g(x)=u2g(x) = u^2.
  • First, differentiate u2u^2: ddu(u2)=2u\frac{d}{du}(u^2) = 2u.
  • Now differentiate u=sin(3x2)u = \sin(3x^2): The derivative of sin(x)\sin(x) is cos(x)\cos(x), so ddxsin(3x2)=cos(3x2)6x\frac{d}{dx}\sin(3x^2) = \cos(3x^2) \cdot 6x (by applying the chain rule again to 3x23x^2).

Now combine everything: g(x)=2sin(3x2)(6xcos(3x2))=12xsin(3x2)cos(3x2)g'(x) = 2\sin(3x^2) \cdot (6x \cos(3x^2)) = 12x \sin(3x^2) \cos(3x^2)

These are the solutions for the first five problems. Would you like detailed explanations for more problems or do you have any specific questions?


Here are five related questions for further understanding:

  1. How is the chain rule different from the product rule?
  2. What would happen if a function includes multiple inner functions? How would the chain rule apply?
  3. Can the chain rule be applied to implicit differentiation problems? How?
  4. How do you differentiate trigonometric functions that have nested functions inside?
  5. In what cases might you need to apply the chain rule more than once?

Tip: Always identify the "inner" and "outer" functions when applying the chain rule. This makes the process clearer and more structured.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Chain Rule
Trigonometric Derivatives
Exponential Functions

Formulas

Chain Rule: d/dx[f(g(x))] = f'(g(x)) * g'(x)
Derivative of sin(x): cos(x)
Derivative of cos(x): -sin(x)
Power Rule: d/dx[x^n] = n*x^(n-1)

Theorems

Chain Rule Theorem
Derivative Rules for Exponential and Trigonometric Functions

Suitable Grade Level

College/University Level