Math Problem Statement

Evaluate the derivative of the following functions.

Solution

You uploaded an image with several calculus problems involving derivatives. Let's go over the solutions for each of them:

1. ddx(x)\frac{d}{dx} (x)

  • The derivative of xx is: ddx(x)=1\frac{d}{dx}(x) = 1

2. y=x3x+9y = x^3 - x + 9

  • Differentiate each term: dydx=3x21\frac{dy}{dx} = 3x^2 - 1

3. y=4x2+5x12y = 4x^2 + 5x - \frac{1}{2}

  • The derivative of a constant like 12-\frac{1}{2} is zero: dydx=8x+5\frac{dy}{dx} = 8x + 5

4. ddx((3x52x3+1)(x423x3+5))\frac{d}{dx} \left( (3x^5 - 2x^3 + 1)(x^4 - \frac{2}{3}x^3 + 5) \right)

  • Use the product rule: ddx(uv)=uv+uv\frac{d}{dx}(uv) = u'v + uv'.
    • Let u=3x52x3+1u = 3x^5 - 2x^3 + 1, so u=15x46x2u' = 15x^4 - 6x^2
    • Let v=x423x3+5v = x^4 - \frac{2}{3}x^3 + 5, so v=4x32x2v' = 4x^3 - 2x^2

The derivative will be: ddx=(15x46x2)(x423x3+5)+(3x52x3+1)(4x32x2)\frac{d}{dx} = (15x^4 - 6x^2)(x^4 - \frac{2}{3}x^3 + 5) + (3x^5 - 2x^3 + 1)(4x^3 - 2x^2) Simplify the expression after multiplying the terms.

5. y=x95x2+16y = x^9 - 5x^2 + 16

  • Differentiate term by term: dydx=9x810x\frac{dy}{dx} = 9x^8 - 10x

6. (This is the same as number 5, so the result will be the same.)

7. y=5x9y = \frac{5}{x^9}

  • Rewrite the function as y=5x9y = 5x^{-9} and differentiate: dydx=45x10\frac{dy}{dx} = -45x^{-10}

8. ddx(23x3+12x2+2x56)\frac{d}{dx} \left( \frac{2}{3}x^3 + \frac{1}{2}x^2 + 2x - \frac{5}{6} \right)

  • Differentiate each term: ddx=2x2+x+2\frac{d}{dx} = 2x^2 + x + 2

9. ddx(5x9)\frac{d}{dx} \left( \frac{5}{x^9} \right)

  • Rewrite as 5x95x^{-9} and differentiate: ddx=45x10\frac{d}{dx} = -45x^{-10}

10. y=8x8+x8y = \frac{8}{x^8} + \frac{x}{8}

  • Rewrite the first term as 8x88x^{-8} and differentiate: dydx=64x9+18\frac{dy}{dx} = -64x^{-9} + \frac{1}{8}

11. ddx(57x756x5+7x1)\frac{d}{dx} \left( \frac{5}{7}x^7 - \frac{5}{6}x^5 + 7x - 1 \right)

  • Differentiate each term: ddx=5x6256x4+7\frac{d}{dx} = 5x^6 - \frac{25}{6}x^4 + 7

12. y=2x3(1+1x3)y = 2x^3 \left( 1 + \frac{1}{x^3} \right)

  • Simplify the function first: y=2x3(x3+1x3)=2(x3+1)y = 2x^3 \cdot \left( \frac{x^3 + 1}{x^3} \right) = 2(x^3 + 1)
  • Now, differentiate: dydx=6x2\frac{dy}{dx} = 6x^2

13. y=x(x3+x)y = \sqrt{x} \left( \sqrt[3]{x} + x \right)

  • First simplify and apply the product rule:
    • Rewrite x\sqrt{x} as x1/2x^{1/2} and x3\sqrt[3]{x} as x1/3x^{1/3} y=x1/2(x1/3+x)y = x^{1/2} \left( x^{1/3} + x \right) Differentiate using the product rule.

14. ddx(1x3)\frac{d}{dx} \left( \frac{1}{\sqrt[3]{x}} \right)

  • Rewrite as x1/3x^{-1/3} and differentiate: ddx=13x4/3\frac{d}{dx} = -\frac{1}{3}x^{-4/3}

15. y=(x23x+1)(x45x2x+9)y = (x^2 - 3x + 1)(x^4 - 5x^2 - x + 9)

  • Use the product rule:
    • Let u=x23x+1u = x^2 - 3x + 1 and v=x45x2x+9v = x^4 - 5x^2 - x + 9
    • Differentiate each and apply the product rule.

Would you like detailed steps for any of these problems, or explanations for any specific rules used?

Further Questions:

  1. Can you explain the product rule in detail with an example?
  2. How do you handle the chain rule when dealing with square roots?
  3. What is the best approach for differentiating polynomials with large exponents?
  4. How can logarithmic differentiation simplify problems like question 12?
  5. Can you clarify the difference between the quotient rule and rewriting as negative exponents?

Tip:

When solving derivatives of products or quotients, it's helpful to first simplify the expressions as much as possible. This can often make the differentiation process much quicker and less error-prone.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Polynomial functions
Product rule
Quotient rule
Chain rule

Formulas

Power Rule: d/dx (x^n) = n * x^(n-1)
Product Rule: d/dx (u * v) = u' * v + u * v'
Quotient Rule: d/dx (u/v) = (u' * v - u * v') / v^2
Chain Rule: d/dx [f(g(x))] = f'(g(x)) * g'(x)

Theorems

Fundamental Theorem of Calculus
Power rule
Product rule
Quotient rule

Suitable Grade Level

Undergraduate calculus or high school AP calculus