Math Problem Statement

show me how u put and solve the values for double diff

Solution

Sure! Let's walk through the process of solving a double derivative (second-order derivative) step by step.

General Steps for Finding a Double Derivative

Let's assume we have a function f(x)f(x) and we want to find its second derivative.

Step 1: Find the first derivative f(x)f'(x)

To find the first derivative, we apply the basic rules of differentiation to the function f(x)f(x). This could include:

  • Power rule: ddxxn=nxn1\frac{d}{dx}x^n = n \cdot x^{n-1}
  • Product rule: ddx(uv)=uv+uv\frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v'
  • Quotient rule: ddx(uv)=vuuvv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}
  • Chain rule: ddxf(g(x))=f(g(x))g(x)\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)

Step 2: Find the second derivative f(x)f''(x)

The second derivative is simply the derivative of the first derivative. This means you will apply the differentiation rules to f(x)f'(x) to obtain f(x)f''(x).

Example

Let’s take an example function: f(x)=3x45x3+2x2+x7f(x) = 3x^4 - 5x^3 + 2x^2 + x - 7

Step 1: Find the first derivative f(x)f'(x)

We differentiate each term:

  • The derivative of 3x43x^4 is 12x312x^3.
  • The derivative of 5x3-5x^3 is 15x2-15x^2.
  • The derivative of 2x22x^2 is 4x4x.
  • The derivative of xx is 11.
  • The derivative of 7-7 is 00.

So the first derivative is: f(x)=12x315x2+4x+1f'(x) = 12x^3 - 15x^2 + 4x + 1

Step 2: Find the second derivative f(x)f''(x)

Now, we differentiate f(x)=12x315x2+4x+1f'(x) = 12x^3 - 15x^2 + 4x + 1:

  • The derivative of 12x312x^3 is 36x236x^2.
  • The derivative of 15x2-15x^2 is 30x-30x.
  • The derivative of 4x4x is 44.
  • The derivative of 11 is 00.

Thus, the second derivative is: f(x)=36x230x+4f''(x) = 36x^2 - 30x + 4

Summary of Results

  • First derivative: f(x)=12x315x2+4x+1f'(x) = 12x^3 - 15x^2 + 4x + 1
  • Second derivative: f(x)=36x230x+4f''(x) = 36x^2 - 30x + 4

Want More Details or Have Questions?

  1. How would the differentiation process change for trigonometric functions like sin(x)\sin(x)?
  2. What happens if there is a composite function involved, such as f(x)=sin(x2)f(x) = \sin(x^2)?
  3. How do implicit functions affect the process of taking derivatives?
  4. Can we apply the chain rule more than once in a single function?
  5. How does the second derivative help in analyzing the concavity of a function?

Tip: When calculating derivatives, be mindful of signs and powers—these are easy spots where errors can occur!

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Second-Order Derivatives

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) = (v * u' - u * v') / v^2
Chain rule: d/dx(f(g(x))) = f'(g(x)) * g'(x)

Theorems

Rules of Differentiation

Suitable Grade Level

Grades 11-12 or early college