What's h'(x)?

Question

If h(x)=f^2 (x)-g^2 (x), f'(x)=-g(x), and g'(x)=f(x), what's h'(x)?

 

Answer: -4f(x)g(x)

 

 

Answer 1

 

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)

 

 

 

  

Answer 2

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)