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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB>Unit 3

Lesson 10: Optional videos

# Chain rule proof

Here we use the formal properties of continuity and differentiability to see why the chain rule is true.

## Want to join the conversation?

• " lim_{Δx->0} (Δy/Δu) * (Δu)(Δx) "

But if Δu=0 (even when Δx ≠ 0 ), you'd be dividing by zero ? This reasoning suggests that the chain rule is true but I don't think it's rigorous enough. •  You are correct. This is the "intuitive" proof. There is a rigorous proof, the chain rule is sound.
To prove the Chain Rule correctly you need to show that if f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then the composite y=f(g(x)) is a differentiable function of x. Since a function is differentiable if and only if it has a derivative at each point in its domain, it must be shown that whenever g is differentiable at xₒ and f is differentiable at g(xₒ) , then the composite is differentiable at xₒ and the derivative of the composite satisfies the equation:
dy/dx, = f'(g(xₒ))·g'(xₒ) (when x=xₒ,)
Good eye!
• Is it true that every type of derivative is actually also a chain rule on top of whatever other type it is? I've always had the sneaking suspicion that this is true and I haven't yet found a counterexample, but in math you need formal proofs. Let me provide an example of what I am talking about, when I take the derivative of f(x) = 3x^2 and get my result of d/dx(x) = 2*3x^(2-1) = 6x, isn't that still a chain rule, but I just didn't have to type out the second part of the chain rule because the derivative of the inside x is just 1, so that would have made the (complete) way to take the derivative using the chain rule and power rule d/dx(3x^2) = [2*3x^(2-1)] * (d/dx(x)) = 6x * 1 = 6x. I get the same result, but it shows that the chain rule still holds for different types of derivatives besides just standard chain rule problems. But does this discovery hold for every derivative? • Should not the function y be differentiable at u(x) and not x? • can't you cancel out du directly, like you would cancel out the 2 in 2/3*1/2= 1/3 for dy/du*du/dx= dy/dx? • At , Sal says d/dx [y(u(x))] = (dy/du) * (du/dx). Shouldn't this be (d [y(u(x))]/du) * (du/dx)? Because Sal is implying that d/dx [f(g(x))] = (d [f(x)]/d [g(x)]) * (d [g(x)])/dx. • So you wrote: "Shouldn't this be (d [y(u(x))]/du) * (du/dx)?"
Try this instead: ( d [y(u(x)]/d[u(x)] ) * ( d[u(x)] / dx )

Notice I included a the whole "u(x)" in place of the lone "u" that you sometimes wrote. By writing just "u", you just used a short hand notation for "u(x)". This is what Sal is doing doing by writing dy/du, instead of: ( d[y(u(x))] / d[u(x)] ) * ( d[u(x)] / dx ). Either way is fine, if you know that y is short for y(u(x)).
• why does f'(g(x)) equal to dy/du, can someone please explain, thanks a lot.☻ • at sal calls the chain rule infamous, just asking but why • By the way, what does dy/du mean? It doesn't make sense to me because u is a function not a variable... • Exesssr is incorrect, Sal is talking about differentiating y(u(x)) with respect to u(x). A function is a dependent variable with only one value for any given value of the independent variable. Therefore, y(u(x)) is a variable dependent on u(x), which in turn is a variable dependent on x. u(x) is not nessesarily equal to x, however, so dy/du /= dy/dx. In fact, dy/du * du/dx = dy/dx, so dy/du only equals dy/dx when du/dx equals 1 (du/dx can only be constantly equal 1 if u=x).
• Why can't I just say that dx, dy and du are infinitesimal changes and hence directly prove the chain rule by multiplication and division of du?
dy/dx = dy/du . du/dx.
Even if 'd' corresponds to a very,very small change, still it is a change in variable and I should be able to do algebraic manipulation? • Derivatives can be defined in two ways, using limits or using infinitesimals. We cannot assume that the infinitesimal change du in dy/du is equal to the infinitesimal change du in du/dx. However, assuming both dy/du and du/dx are differentiable, the standard part of dy/du and du/dx must be constant for ANY infinitesimal du or dx. In order to be differentiable, the standard part of dy/du and du/dx must also be defined, and therefore du in du/dx must be infinitesimal when dx is infinitesimal. Therefore, we can define du in dy/du to be the infinitesimal change of du for a given dx, knowing that it will be the same for any other du. From there, we can algebraicly solve dy/du * du/dx to get dy/dx. Sal essentially did the same thing that I'm doing here, except he used limits instead of infinitesimals and worked in reverse order. 