Math Problem Statement

Find dy/dx if y = 5 / sqrt(x^3 - 2x)

Solution

The problem asks you to find dydx\frac{dy}{dx} for the function:

y=5x32xy = \frac{5}{\sqrt{x^3 - 2x}}

Let's solve this step-by-step.

Step 1: Rewrite the function for easier differentiation

The given function can be rewritten as: y=5(x32x)12y = 5 \cdot (x^3 - 2x)^{-\frac{1}{2}}

Step 2: Apply the chain rule

We'll need to apply the chain rule and the power rule to differentiate.

  1. The outer function is u12u^{-\frac{1}{2}}, where u=x32xu = x^3 - 2x.

    • The derivative of u12u^{-\frac{1}{2}} is: 12u32dudx-\frac{1}{2} u^{-\frac{3}{2}} \cdot \frac{du}{dx}
  2. Now, calculate dudx\frac{du}{dx}, where u=x32xu = x^3 - 2x: dudx=3x22\frac{du}{dx} = 3x^2 - 2

Step 3: Combine everything

Now apply the chain rule:

dydx=5(12)(x32x)32(3x22)\frac{dy}{dx} = 5 \cdot \left( -\frac{1}{2} \right) \cdot (x^3 - 2x)^{-\frac{3}{2}} \cdot (3x^2 - 2)

Simplify:

dydx=5(3x22)2(x32x)32\frac{dy}{dx} = -\frac{5(3x^2 - 2)}{2(x^3 - 2x)^{\frac{3}{2}}}

Step 4: Compare with the options

The correct option should match this expression. Let's compare the final form with the multiple-choice options:

The correct answer is:

a\boxed{a}

Let me know if you'd like further details on this solution!

Here are five related questions to explore this further:

  1. How would the process change if the exponent were different, such as (x32x)34(x^3 - 2x)^{-\frac{3}{4}}?
  2. What would happen if we had a different coefficient, say y=7x32xy = \frac{7}{\sqrt{x^3 - 2x}}?
  3. Can you apply this method to a similar function like y=10x2+3xy = \frac{10}{\sqrt{x^2 + 3x}}?
  4. How does the chain rule help in simplifying complex derivatives?
  5. What is the significance of the negative exponent when differentiating functions like (x32x)12(x^3 - 2x)^{-\frac{1}{2}}?

Tip:

Remember, always simplify the function as much as possible before differentiating. This can make the process easier and reduce errors.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Chain Rule
Power Rule
Algebraic Functions

Formulas

dy/dx = (dy/du) * (du/dx)
dy/dx = -5(3x^2 - 2) / [2(x^3 - 2x)^(3/2)]

Theorems

Chain Rule
Power Rule

Suitable Grade Level

Grade 11-12, Calculus