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Simplify the function before differentiating

Consider the function f(x)=x2−1x+1f(x) = \dfrac{x^2 - 1}{x + 1}, where x≠−1x \neq -1.

Simplify f(x)=x2−1x+1f(x) = \dfrac{x^2-1}{x+1} into an equivalent expression that no longer has x+1x+1 in the denominator. What is the simplified form of f(x)f(x)?

Hint

Factor the numerator as a difference of squares, then cancel the common factor with the denominator.

Step-by-step solution

Factor the numerator: x2−1=(x−1)(x+1)x^2 - 1 = (x-1)(x+1). So f(x)=(x−1)(x+1)x+1f(x) = \dfrac{(x-1)(x+1)}{x+1}.

Cancel the common factor (x+1)(x+1) (valid since x≠−1x \neq -1) to get f(x)=x−1f(x) = x - 1.

Second derivative

Using the simplified form of f(x)f(x) from part (a), find f′′(x)f''(x).

Hint

Differentiate the simplified linear function once to get f′(x)f'(x), then differentiate again.

Step-by-step solution

Since f(x)=x−1f(x) = x - 1, differentiate once: f′(x)=1f'(x) = 1.

Differentiate again: since f′(x)=1f'(x) = 1 is constant, f′′(x)=0f''(x) = 0.

Third derivative

Find f′′′(x)f'''(x).

Hint

Differentiate the result of f′′(x)f''(x) one more time.

Step-by-step solution

From part (b), f′′(x)=0f''(x) = 0 for all x≠−1x \neq -1.

Differentiating a constant (zero) function gives f′′′(x)=0f'''(x) = 0.

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