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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

577.0. "a Natural log question" by ARGUS::COOK (Hackers Anonymous) Fri Sep 12 1986 18:43

    
    
      Here's a good old Calculus question:
    
      f(x) = ln square root(7-2X**2)   <- 2x squared
    
      find f'(x).
    
        ----------------------
    
     now if f(x) = ln x   then f'(x) = 1/x
    
      
    
      How do you get rid of the square root??
    
    
                                    Peter
                                         
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577.1answerGALLO::EKLUNDDave EklundFri Sep 12 1986 20:567
    	Use the identity:
    
    	ln (sqrt (g(x)) ) == 1/2 * ln (g(x))
    
    	This eliminates the square root.  Chaining then makes the problem
    trivial.
    
577.2CLT::GILBERTeager like a childSat Sep 13 1986 03:5018
    Or, to be unsuave about this, we can apply the chain rule directly:
    
    f(x) = ln (sqrt(7-2x^2))
    
	     d
    f'(x) = -- (sqrt(7-2x^2)) / sqrt(7-2x^2)
	    dx

	    1         -1/2  d
	  = - (7-2x^2)     -- (7-2x^2) / sqrt(7-2x^2)
	    2              dx

	    1         -1/2
	  = - (7-2x^2)     (-4x) / sqrt(7-2x^2)
	    2

	    
	  = -2x / (7-2x^2)