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Second and third derivatives via the power rule

Let f(x)=1x−2f(x) = \dfrac{1}{x-2}.

Rewrite f(x)=1x−2f(x) = \dfrac{1}{x-2} as a power of (x−2)(x-2), then find f′′(x)f''(x) and f′′′(x)f'''(x).

Hint

Write f(x)=(x−2)−1f(x) = (x-2)^{-1} and apply the power rule repeatedly, treating (x−2)(x-2) as the variable being raised to a power (its derivative is just 11).

Step-by-step solution

Rewrite f(x)=(x−2)−1f(x) = (x-2)^{-1}. Differentiating once gives f′(x)=−1(x−2)−2⋅1=−(x−2)−2f'(x) = -1(x-2)^{-2}\cdot 1 = -(x-2)^{-2}, since the derivative of (x−2)(x-2) is 11.

Differentiate f′(x)=−(x−2)−2f'(x) = -(x-2)^{-2} again: f′′(x)=−(−2)(x−2)−3=2(x−2)−3=2(x−2)3f''(x) = -(-2)(x-2)^{-3} = 2(x-2)^{-3} = \dfrac{2}{(x-2)^3}.

Differentiate once more: f′′′(x)=2(−3)(x−2)−4=−6(x−2)−4=−6(x−2)4f'''(x) = 2(-3)(x-2)^{-4} = -6(x-2)^{-4} = \dfrac{-6}{(x-2)^4}.

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