Thread: Proof of the chain rule View Single Post
 P: 19 how about this one? let u(x) be a differentiable fuction in [a,b] with values in [a',b'] and y=f(x) a differentiable fuction in [a',b']. Let Δx be a randomly picked difference x2-x1. That causes a change Δu on u(x), while Δu causes a change on y=f(u). We have Δu=(u'(x)+n1)*Δx. You can easily verify by looking at the graph that the line connection the points (x1,u(x1)) and (x2,u(x2)) has a slope equal to the value of the derivative of u on x1 plus a number n1 to compensate for the fact that Δx isn't zero(and thus this line isn't the tanget on x1). The same applys to Δy=((f'(u)+n2)*Δu. When Δx-->0 , n1,n2-->0 Δx/Δy=(f;(u)+n2)*(u'(x)+n1) We calculate the limit of the fraction when Δx-->0 and it is equal to f'(u)*u'(x)=(dy/du)*(du/dx) system has gone crazy and wont show the math symbols, sorry for the formating It's the proof from Louis Brand's book "Advanced Calculus", paragraph 52- The chain rule