How do you prove the chain rule?

Question

Charles Anctil · Accepted Answer

see below

i'm gonna make this up on the spot so mea culpa if it's messy but I find Taylor expansions really useful in these sorts of situations if we start with function# f( g(x))# and by definition we have #(df)/(dx) = lim_(h to 0) (f(g(x+h)) - f(g(x)))/(h)# Using Taylor, we're gonna expand the first bit as follows: # g(x+h) = g(x) + h g'(x) + O(h^2)# So to clarify, we have #(df)/(dx) = lim_(h to 0) (f color(red)(( g(x) + h g'(x) + O(h^2))) - f(g(x)))/(h)# now to simplify a little we set: #eta(x) = g'(x) + O(h)# #(df)/(dx) = lim_(h to 0) (f ( g(x) + h eta (x) ) - f(g(x)))/(h) qquad square# And now we're gonna expand #f ( g(x) + eta (x) )# by the same process #f ( g(x) + h eta (x) ) = f(g(x)) + h eta (x) f'(g(x)) + O(h^2)# #= f(g(x)) + h ( g'(x) + O(h)) f'(g(x)) + O(h^2)# #= f(g(x)) + h f'(g(x)) g'(x) + O(h^2)# We can put that in #square# #(df)/(dx) = lim_(h to 0) ( f(g(x)) + h f'(g(x)) g'(x) + O(h^2)- f(g(x)))/(h)# #(df)/(dx) = lim_(h to 0) ( h f'(g(x)) g'(x) + O(h^2))/(h)# #= lim_(h to 0) f'(g(x)) g'(x) + O(h)# #= f'(g(x)) g'(x) #

Charles Anctil · Answer

undefined To prove the chain rule, start with the definition of the derivative. Then, apply the limit definition of the derivative to the composition of two functions, $ f(g(x)) $. Using algebraic manipulation and the properties of limits, simplify the expression until it matches the form of the chain rule. Finally, show that as $ h $ approaches zero, the expression converges to the chain rule formula, which states that the derivative of $ f(g(x)) $ with respect to $ x $ is equal to the derivative of $ f $ with respect to $ g $ multiplied by the derivative of $ g $ with respect to $ x $.