Math Problem Statement

Find f''(x) for the function f(x) = x / (x + 1).

Solution

To find the second derivative f(x)f''(x) of the function given:

f(x)=xx+1f(x) = \frac{x}{x + 1}

we'll go through the following steps:

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

We use the quotient rule for differentiation, which states that if f(x)=g(x)h(x)f(x) = \frac{g(x)}{h(x)}, then f(x)=g(x)h(x)g(x)h(x)[h(x)]2f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}

For our function:

  • g(x)=xg(x) = x and g(x)=1g'(x) = 1
  • h(x)=x+1h(x) = x + 1 and h(x)=1h'(x) = 1

Thus, f(x)=(1)(x+1)(x)(1)(x+1)2=x+1x(x+1)2=1(x+1)2f'(x) = \frac{(1)(x+1) - (x)(1)}{(x + 1)^2} = \frac{x + 1 - x}{(x + 1)^2} = \frac{1}{(x + 1)^2}

So, the first derivative is: f(x)=1(x+1)2f'(x) = \frac{1}{(x + 1)^2}

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

Now, we differentiate f(x)=1(x+1)2f'(x) = \frac{1}{(x + 1)^2} again. We can rewrite it as: f(x)=(x+1)2f'(x) = (x + 1)^{-2}

Using the chain rule: f(x)=2(x+1)31=2(x+1)3f''(x) = -2(x + 1)^{-3} \cdot 1 = -\frac{2}{(x + 1)^3}

Final Answer

The second derivative of the function is: f(x)=2(x+1)3f''(x) = -\frac{2}{(x + 1)^3}

Would you like more details or have any questions?

Here are some related questions to consider:

  1. How does the quotient rule work for finding derivatives?
  2. Why do we use the chain rule when differentiating (x+1)2(x + 1)^{-2}?
  3. How would the derivatives change if f(x)f(x) had a different denominator?
  4. What is the significance of the sign of f(x)f''(x) in terms of concavity?
  5. Can f(x)=xx+1f(x) = \frac{x}{x + 1} have any inflection points?

Tip: When differentiating functions with complex denominators, rewriting them in exponent form can simplify the process, especially for higher derivatives.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Quotient Rule
Chain Rule

Formulas

Quotient Rule: (g(x)/h(x))' = (g'(x)h(x) - g(x)h'(x)) / [h(x)]^2
Chain Rule: d/dx [f(g(x))] = f'(g(x)) * g'(x)

Theorems

Quotient Rule for Differentiation
Chain Rule for Differentiation

Suitable Grade Level

Grades 11-12