Math Problem Statement

Find the first and second derivatives of y = (2x + 3)(5x^2 - 4x).

Solution

We are given the function:

y=(2x+3)(5x24x)y = (2x + 3)(5x^2 - 4x)

We are tasked with finding the first and second derivatives of this function.

Step 1: Expand the function

First, let's expand the product:

y=(2x+3)(5x24x)y = (2x + 3)(5x^2 - 4x)

Using distribution:

y=(2x)(5x24x)+(3)(5x24x)y = (2x)(5x^2 - 4x) + (3)(5x^2 - 4x)

y=2x(5x2)2x(4x)+3(5x2)3(4x)y = 2x(5x^2) - 2x(4x) + 3(5x^2) - 3(4x)

y=10x38x2+15x212xy = 10x^3 - 8x^2 + 15x^2 - 12x

Combine like terms:

y=10x3+7x212xy = 10x^3 + 7x^2 - 12x

Step 2: First derivative (dy/dx)

Now, differentiate term by term:

dydx=ddx(10x3)+ddx(7x2)+ddx(12x)\frac{dy}{dx} = \frac{d}{dx}(10x^3) + \frac{d}{dx}(7x^2) + \frac{d}{dx}(-12x)

dydx=30x2+14x12\frac{dy}{dx} = 30x^2 + 14x - 12

Step 3: Second derivative (d²y/dx²)

Now, differentiate the first derivative:

d2ydx2=ddx(30x2+14x12)\frac{d^2y}{dx^2} = \frac{d}{dx}(30x^2 + 14x - 12)

d2ydx2=60x+14\frac{d^2y}{dx^2} = 60x + 14

Summary

  • First derivative: dydx=30x2+14x12\frac{dy}{dx} = 30x^2 + 14x - 12
  • Second derivative: d2ydx2=60x+14\frac{d^2y}{dx^2} = 60x + 14

Would you like further clarification or a breakdown of any of the steps?

Here are some related questions to explore further:

  1. How do we apply the product rule for derivatives?
  2. What is the importance of the second derivative in curve analysis?
  3. How do we interpret the first derivative in terms of function behavior?
  4. Can we derive higher-order derivatives for this function?
  5. What is the geometric significance of the second derivative?

Tip: Always simplify the function first before differentiating, as it makes the process easier and less error-prone.

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Math Problem Analysis

Mathematical Concepts

Derivatives
Algebra
Polynomial Differentiation

Formulas

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

Theorems

Product Rule for Derivatives
Power Rule for Derivatives

Suitable Grade Level

Grades 11-12 (Calculus)