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Find the derivative of r(x) r(x) . \newliner(x)=ln⁡(1x) r(x) = \ln\left(\frac{1}{x}\right) \newliner′(x)= r'(x) = ______

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Q. Find the derivative of r(x) r(x) . \newliner(x)=ln⁡(1x) r(x) = \ln\left(\frac{1}{x}\right) \newliner′(x)= r'(x) = ______
  1. Recognize logarithm properties: Recognize that r(x)=ln⁡(1x)r(x) = \ln(\frac{1}{x}) can be rewritten using the properties of logarithms as r(x)=−ln⁡(x)r(x) = -\ln(x).
  2. Find derivative of r(x)r(x): Now that we have r(x)=−ln⁡(x)r(x) = -\ln(x), we can find the derivative of r(x)r(x) with respect to xx. The derivative of ln⁡(x)\ln(x) with respect to xx is 1x\frac{1}{x}.
  3. Apply chain rule: Since r(x)=−ln⁡(x)r(x) = -\ln(x), the derivative r′(x)r'(x) is the derivative of −ln⁡(x)-\ln(x), which is −1-1 times the derivative of ln⁡(x)\ln(x). Therefore, r′(x)=−1×(1/x)r'(x) = -1 \times (1/x).
  4. Simplify the expression: Simplify the expression for r′(x)r'(x) to get the final answer. r′(x)=−1xr'(x) = -\frac{1}{x}.

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