How do we derive the product rule?
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"What is Boyle's law?"
Let #f(x)=g(x)h(x)#.
Now from definition #(df)/(dx)=Lt_(deltax->o)(f(x+deltax)-f(x))/(deltax)#
also #(dg)/(dx)=Lt_(deltax->o)(g(x+deltax)-g(x))/(deltax)#
and #(dh)/(dx)=Lt_(deltax->o)(h(x+deltax)-h(x))/(deltax)#
As #f(x)=g(x)h(x)# and #f(x+deltax)=g(x+deltax)h(x+deltax)#
#(df)/(dx)=Lt_(deltax->o)(f(x+deltax)-f(x))/(deltax)#
= #Lt_(deltax->o)(g(x+deltax)h(x+deltax)-g(x)h(x))/(deltax)#
= #Lt_(deltax->o)(g(x+deltax)h(x+deltax)color(red)(-g(x)h(x+deltax)+g(x)h(x+deltax))-g(x)h(x))/(deltax)#
= #Lt_(deltax->o)(g(x+deltax)h(x+deltax)-g(x)h(x+deltax))/(deltax)+Lt_(deltax->o)(g(x)h(x+deltax)-g(x)h(x))/(deltax)#
= #Lt_(deltax->o)[h(x+deltax)(g(x+deltax)-g(x))/(deltax)]+g(x)Lt_(deltax->o)[(h(x+deltax)-h(x))/(deltax)]#
= #h(x)(dg)/(dx)+g(x)(dh)/(dx)#
I use the product rule in the form:
#d/dx(f(x)g(x)) = f'(x)g(x)+f(x)g'(x)#
(Because multiplication and addition are commutative, it can also be written in other orders.)
Proof:
Suppose that #f# and #g# are differentiable at #x#.
Let #P(x) = f(x)g(x)#.
(We will show that #P'(x)= f'(x)g(x)+f(x)g'(x)#.)
Then
#P'(x) = lim_(hrarr0)(P(x+h)-P(x))/h#
# = lim_(hrarr0)(f(x+h)g(x+h) - f(x)g(x))/h#
Now we will add #0# in the numerator. We'll use #-f(x)g(x+h)+f(x)g(x+h)#. (See note (1) below.)
# = lim_(hrarr0)(f(x+h)g(x+h) -f(x)g(x+h)+f(x)g(x+h) - f(x)g(x))/h#
Now regroup:
# = lim_(hrarr0)((f(x+h) -f(x))g(x+h)+f(x)(g(x+h) - g(x)))/h#
Rewrite as the sum of two ratios and factor.
# = lim_(hrarr0)((f(x+h) -f(x))g(x+h))/h + (f(x)(g(x+h) - g(x)))/h#
# = lim_(hrarr0)((f(x+h) -f(x)))/hg(x+h) + lim_(hrarr0)f(x)((g(x+h) - g(x)))/h#
We now have limits of 4 things to evaluate.
# lim_(hrarr0)((f(x+h) -f(x)))/h = f'(x)#
#lim_(hrarr0)g(x+h)=g(x)# (see note (2) below).
#lim_(hrarr0)f(x) = f(x)#
# lim_(hrarr0)((g(x+h) -g(x)))/h = g'(x)#
Finally we write the total limit:
# = f'(x)g(x)+f(x)g'(x)#
That is:
#P(x) = f(x)g(x)# has derivative
#P'(x) = f'(x)g(x)+f(x)g'(x)#
Notes
(1) To prove the product rule in a different order we would add #0# in the form #-f(x+h)g(x)+f(x+h)g(x)# and we'd end with #f(x)g'(x)+f'(x)g(x)#
(2) By hypothesis #g# is differentiable at #x#, so #g# is continuous at #x# and therefore #lim_(hrarr0)g(x+h)=g(x)#.
