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Find the derivative of (x^2 + 1)/(x - 3)

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Answer
f′(x)=x2−6x−1(x−3)2\boxed{f'(x) = \frac{x^2 - 6x - 1}{(x - 3)^2}}

We can verify this result using the quotient rule formula u′v−uv′v2\frac{u'v - uv'}{v^2}, where u=x2+1u = x^2 + 1 and v=x−3v = x - 3.

Problem: Find the derivative of f(x)=x2+1x−3f(x) = \frac{x^2 + 1}{x - 3}.

Steps

  1. Apply the quotient rule
    f′(x)=(2x)(x−3)−(x2+1)(1)(x−3)2f'(x) = \frac{(2x)(x - 3) - (x^2 + 1)(1)}{(x - 3)^2}
  2. Expand the numerator
    f′(x)=(2x2−6x)−(x2+1)(x−3)2f'(x) = \frac{(2x^2 - 6x) - (x^2 + 1)}{(x - 3)^2}
  3. Combine like terms in the numerator
    f′(x)=2x2−6x−x2−1(x−3)2f'(x) = \frac{2x^2 - 6x - x^2 - 1}{(x - 3)^2}
  4. Simplify the expression
    f′(x)=x2−6x−1(x−3)2f'(x) = \frac{x^2 - 6x - 1}{(x - 3)^2}

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