text: # What is the slope of the tangent line of 1/(e^y-e^x) = C , where C is an arbitrary constant, at (-2,1)?
  
  Aug 6, 2016
  
  The slope is $\frac{1}{e} ^ 3$.
  
  #### Explanation:
  
  This will be easier rewritten as:
  
  ${\left({e}^{y} - {e}^{x}\right)}^{-} 1 = C$
  
  Now, we can use implicit differentiation (remember that each occurrence of $y$ will put the chain rule in effect):
  
  $- {\left({e}^{y} - {e}^{x}\right)}^{-} 2 \cdot \frac{d}{\mathrm{dx}} \left({e}^{y} - {e}^{x}\right) = 0$
  
  $- \frac{1}{{e}^{y} - {e}^{x}} ^ 2 \left({e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{x}\right) = 0$
  
  $\frac{{e}^{x} - {e}^{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{e}^{y} - {e}^{x}} ^ 2 = 0$
  
  So, we want to solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$ when $\left(x , y\right) = \left(- 2 , 1\right)$.
  
  $\frac{{e}^{-} 2 - {e}^{1} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{e}^{1} - {e}^{-} 2} ^ 2 = 0$
  
  Cross-multiplying:
  
  $\frac{1}{e} ^ 2 - e \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$
  
  $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{e} ^ 3$
explanation: The document contains a math question and its corresponding answer. The question asks for the slope of the tangent line of the equation \( \frac{1}{e^y - e^x} = C \) at the point \((-2, 1)\). The answer provided is \( \frac{1}{e^3} \).
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # What is the slope of the tangent line of 1/(e^y-e^x) = C , where C is an arbitrary constant, at (-2,1)?
  
  Aug 6, 2016
  
  The slope is $\frac{1}{e} ^ 3$.
  
  #### Explanation:
  
  This will be easier rewritten as:
  
  ${\left({e}^{y} - {e}^{x}\right)}^{-} 1 = C$
  
  Now, we can use implicit differentiation (remember that each occurrence of $y$ will put the chain rule in effect):
  
  $- {\left({e}^{y} - {e}^{x}\right)}^{-} 2 \cdot \frac{d}{\mathrm{dx}} \left({e}^{y} - {e}^{x}\right) = 0$
  
  $- \frac{1}{{e}^{y} - {e}^{x}} ^ 2 \left({e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{x}\right) = 0$
  
  $\frac{{e}^{x} - {e}^{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{e}^{y} - {e}^{x}} ^ 2 = 0$
  
  So, we want to solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$ when $\left(x , y\right) = \left(- 2 , 1\right)$.
  
  $\frac{{e}^{-} 2 - {e}^{1} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{{e}^{1} - {e}^{-} 2} ^ 2 = 0$
  
  Cross-multiplying:
  
  $\frac{1}{e} ^ 2 - e \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$
  
  $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{e} ^ 3$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and its corresponding answer. The question asks for the slope of the tangent line of the equation \( \frac{1}{e^y - e^x} = C \) at the point \((-2, 1)\). The answer provided is \( \frac{1}{e^3} \).
  
  Label: question-and-answer