Partial Derivatives > partial derivative of (z-ln(y)/(z-x))
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∂∂x (z−ln(y)z−x )=z−ln(y)(z−x)2 
 
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∂ ∂ x (z−ln(y)z−x )=z−ln(y)(z−x)2 
  • steps
    • ∂ ∂ x (z−ln(y)z−x )
  • Treat z, y as constants
  • Take the constant out:    (a·f)′=a·f ′
    =(z−ln(y))∂ ∂ x (1z−x )
  • Apply exponent rule:    1a =a−1
    =(z−ln(y))∂ ∂ x ((z−x)−1)
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    Apply the chain rule:    −1(z−x)2 ∂ ∂ x (z−x)
    =−1(z−x)2 ∂ ∂ x (z−x)
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    ∂ ∂ x (z−x)=−1
    =(z−ln(y))(−1(z−x)2 (−1))
  • show steps
    Simplify (z−ln(y))(−1(z−x)2 (−1)):    z−ln(y)(z−x)2 
    =z−ln(y)(z−x)2 

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