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What's h'(x)?

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2 Answers

h(x) = f2(x) - g2(x)
h'(x) = [f2(x) - g2(x)]' = [f2(x)]' - [g2(x)]'
 
We need to use the chain rule to find [f2(x)]' and [g2(x)]', that is, we need to take the derivative of (something)2, which is 2(something), and then multiply that by the derivative of that something.
 
[f2(x)]' - [g2(x)]' = 2f(x)f'(x) - 2g(x)g'(x)
                         = 2f(x)[-g(x)] - 2g(x)[f(x)]
                         = -2f(x)g(x) - 2f(x)g(x)
                         = -4f(x)g(x)
            

Comments

Comment

 
Given :  h( x) = f2(x) - g2 (X) =
    
             f'(X) = g( x)
 
            g' ( x) = f ( X)
 
  Slove for  h‘( x) = ?
 
              h' ( x ) = 2 f(x)(- f ‘(x)) - 2 g(x) g‘(x) =
                        
                         = 2 g‘ (x)(- g(x)) - 2 g( x) g'(x) =0
                         = -4 g' (x) g(x)
                          or  -4 f(x)g(x)