Math Problem Statement

Given the function y, equals, 3, cosine, left parenthesis, square root of, 6, minus, 4, x, cubed, end square root, right parenthesis, commay=3cos(6−4x3​), find start fraction, d, y, divided by, d, x, end fraction, .dxdy

Solution

To find the derivative of the given function y=3cos(64x3)y = 3\cos(\sqrt{6 - 4x^3}) with respect to xx, we apply the chain rule. Let's go step by step:

Step 1: Outer function derivative

The outer function is y=3cos(u)y = 3\cos(u), where u=64x3u = \sqrt{6 - 4x^3}. The derivative of 3cos(u)3\cos(u) with respect to uu is: ddu[3cos(u)]=3sin(u).\frac{d}{du}[3\cos(u)] = -3\sin(u).

Step 2: Middle function derivative

The middle function is u=64x3u = \sqrt{6 - 4x^3}, which can be rewritten as u=(64x3)1/2u = (6 - 4x^3)^{1/2}. Its derivative with respect to xx is: [ \frac{d}{dx}\left[(6 - 4x^3)^{1/2}\right] = \frac{1}{2}(6 - 4x^3)^{-1/2} \cdot \frac{d}{dx}[6 - 4x^3]. ]

Step 3: Inner function derivative

The inner function is 64x36 - 4x^3. Its derivative with respect to xx is: ddx[64x3]=12x2.\frac{d}{dx}[6 - 4x^3] = -12x^2.

Step 4: Combine the results

Now, substitute the results step by step:

  1. From Step 3: Substitute ddx[64x3]=12x2\frac{d}{dx}[6 - 4x^3] = -12x^2 into Step 2: ddx[u]=12(64x3)1/2(12x2).\frac{d}{dx}[u] = \frac{1}{2}(6 - 4x^3)^{-1/2} \cdot (-12x^2). Simplify: ddx[u]=6x264x3.\frac{d}{dx}[u] = \frac{-6x^2}{\sqrt{6 - 4x^3}}.

  2. From Step 1: Substitute u=64x3u = \sqrt{6 - 4x^3} and dudx=6x264x3\frac{du}{dx} = \frac{-6x^2}{\sqrt{6 - 4x^3}}: dydx=3sin(u)dudx.\frac{dy}{dx} = -3\sin(u) \cdot \frac{du}{dx}. Substitute u=64x3u = \sqrt{6 - 4x^3} and dudx=6x264x3\frac{du}{dx} = \frac{-6x^2}{\sqrt{6 - 4x^3}}: dydx=3sin(64x3)6x264x3.\frac{dy}{dx} = -3\sin(\sqrt{6 - 4x^3}) \cdot \frac{-6x^2}{\sqrt{6 - 4x^3}}.

Step 5: Final simplification

Simplify the expression: dydx=18x2sin(64x3)64x3.\frac{dy}{dx} = \frac{18x^2\sin(\sqrt{6 - 4x^3})}{\sqrt{6 - 4x^3}}.

Final Answer:

dydx=18x2sin(64x3)64x3.\frac{dy}{dx} = \frac{18x^2\sin(\sqrt{6 - 4x^3})}{\sqrt{6 - 4x^3}}.

Would you like further clarification or examples?


Related Questions:

  1. How does the chain rule work in similar problems involving trigonometric functions?
  2. What is the interpretation of this derivative in a physical or geometric context?
  3. How can we simplify sin(64x3)\sin(\sqrt{6 - 4x^3}) for small values of xx?
  4. How does the derivative behave as x0x \to 0?
  5. Can you apply the product rule alongside the chain rule in more complex expressions?

Tip:

Always rewrite expressions in their simplest forms before differentiating to avoid unnecessary complexity.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Trigonometric Functions

Formulas

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

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12 or Introductory College Calculus