Proof of the Product Rule

Key Questions

• How I do I prove the Product Rule for derivatives?

  All we need to do is use the definition of the derivative alongside a simple algebraic trick.

  First, recall the the the product `fg` of the functions `f` and `g` is defined as `(fg)(x)=f(x)g(x)`. Therefore, it's derivative is

  `(fg)^(prime)(x) = lim_(h to 0) ((fg)(x+h)-(fg)(x))/(h) = lim_(h to 0) (f(x+h)g(x+h)-f(x)g(x))/(h)`

  Now, note that the expression above is the same as

  `lim_(h to 0) (f(x+h)g(x+h)+0-f(x)g(x))/(h)`

  Wich we can rewrite, taking into account that `f(x+h)g(x)-f(x+h)g(x)=0`, as:

  `lim_(h to 0) 1/h [f(x+h)g(x+h)+(f(x+h)g(x)-f(x+h)g(x))-f(x)g(x)] = lim_(h to 0) 1/h(f(x+h)[g(x+h)-g(x)]+g(x)[f(x+h)-f(x)])`

  Using the property that the limit of a sum is the sum of the limits, we get:

  `lim_(h to 0) f(x+h)(g(x+h)-g(x))/(h) + lim_(h to 0)g(x)(f(x+h)-f(x))/(h)`

  Wich give us the product rule

  `(fg)^(prime)(x) = f(x)g^(prime)(x)+g(x)f^(prime)(x),`

  since:

  `lim_(h to 0) f(x+h) = f(x),`
  `lim_(h to 0)(g(x+h)-g(x))/(h) = g^(prime)(x),`
  `lim_(h to 0) g(x)=g(x),`
  `lim_(h to 0) (f(x+h)-f(x))/(h) = f^(prime)(x)`

  Vinicius M. G. Silveira · 1 · Feb 1 2015

• How I do I prove the Product Rule for derivatives?