Math Problem Statement

find the derivative

Find the derivative of y = x√x.

Solution

The given function is y=xxy = x \sqrt{x}. To find the derivative, we'll use the product rule along with the rules of differentiation.

Step-by-Step Solution

  1. Rewrite the function: y=xx1/2y = x \cdot x^{1/2}

  2. Apply the product rule: The product rule states that: ddx(uv)=uv+uv\frac{d}{dx}(u \cdot v) = u'v + uv' Here, u=xu = x and v=x1/2v = x^{1/2}.

  3. Differentiate each term:

    • u=xu = x gives u=1u' = 1,
    • v=x1/2v = x^{1/2} gives v=12x1/2v' = \frac{1}{2}x^{-1/2} (using the power rule).
  4. Substitute into the product rule: dydx=uv+uv=(1)(x1/2)+(x)(12x1/2)\frac{dy}{dx} = u'v + uv' = (1)(x^{1/2}) + (x)\left(\frac{1}{2}x^{-1/2}\right)

  5. Simplify the expression:

    • First term: x1/2x^{1/2},
    • Second term: x12x1/2=12x1/2x \cdot \frac{1}{2}x^{-1/2} = \frac{1}{2}x^{1/2}.

    Add the terms: dydx=x1/2+12x1/2\frac{dy}{dx} = x^{1/2} + \frac{1}{2}x^{1/2}

  6. Combine like terms: dydx=22x1/2+12x1/2=32x1/2\frac{dy}{dx} = \frac{2}{2}x^{1/2} + \frac{1}{2}x^{1/2} = \frac{3}{2}x^{1/2}

Final Answer:

dydx=32x\frac{dy}{dx} = \frac{3}{2}\sqrt{x}

Would you like further details or examples on this type of differentiation?


Questions to explore further:

  1. How does the product rule work for more complex functions?
  2. Can this derivative be applied to real-world problems involving roots?
  3. What happens if x<0x < 0 in this context (domain issues)?
  4. How does this derivative change when combined with other operations?
  5. What are other scenarios where roots and powers are involved in differentiation?

Tip: Always simplify roots and exponents to their fractional forms for easier differentiation.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Exponents and Roots
Product Rule

Formulas

Product Rule: d(uv)/dx = u'v + uv'
Power Rule: d(x^n)/dx = nx^(n-1)

Theorems

Product Rule of Differentiation
Power Rule for Exponents

Suitable Grade Level

Grades 11-12