Let's see, I am going to factor out an F of X plus H. Gonna look at this part, this part of the expression. So the first thing I'm gonna do is I'm gonna look at, I'm So all of this is going to be equal to, it's all going to be equal to the limit as H approaches zero. Well to keep going, let's just keep exploring this expression. And at any point you get inspired, I encourage you to pause this video. Subtracted the same thing, but now this thing can be manipulated in interesting algebraic ways to get us to what we all love about the product rule. Now I haven't changed the value, I just added and I don't change the value of this expression. So if we have plus, actually, let me change this, minus F of X plus H, G of X, I can't just subtract, if I subtract it I've got to add it too, so Subtract at the same term here, "I can begin toĪlgebraically manipulate it "and get it to what we all know "as the classic product rule." So what do I add and subtract here? Well let me give you a clue. I'm assuming somebody wasįumbling with it long enough that said, "Oh wait, wait. And what I'm about to show you, I guess you could view itĪs a little bit of a trick. I don't know how to evaluate this limit, there doesn't seem to beĪnything obvious to do. Written it write now, it doesn't seem easy toĪlgebraically manipulate. Now why did I put thisīig, awkward space here? Because just the way I've So you have F of X plus H, G of X plus H minus F of X, G of X, all of that over H. Of applying it to F of X, I applied it to F of X times G of X. All I did so far is I justĪpplied the definition of the derivative, instead So if I just, if I evaluate this at X, this is gonna be minus F of X, G of X. And I'm gonna put aīig, awkward space here and you're gonna see why in a second. That I'm gonna subtract this thing evaluated F of X. So that's going to be F of X plus H, G of X plus H and from I'm gonna write a big, it's gonna be a big rational expression, in the denominator I'm gonna have an H. Well if we just apply theĭefinition of a derivative, that means I'm gonna take the limit as H approaches zero, and the denominator I'm gonna have at H, and the denominator, And if I can come up withĪ simple thing for this, that essentially is the product rule. I want to find the derivative with respect to X, not just of F of X, but the product of two functions, F of X times G of X. Tangent line and all of that, but now I want to do something a little bit more interesting. If we want to think of it visually, this is the slope of the The derivative of it, by definition, by definition, the derivative of F of X is the limit as H approaches zero, of F of X plus H minus F of X, all of that over, all of that over H. So if I have the function F of X, and if I wanted to take So let's just start with ourĭefinition of a derivative. Hope to do in this video is give you a satisfying Having said that, YES, you can use implicit and logarithmic differentiation to do an alternative proof: I think you do understand Sal's (AKA the most common) proof of the product rule. It is this type of insight and intuition, that being, the ability to leverage the rules of mathematics creatively that produces much of the beauty in math. The ability to see that such a device can work is the key to many proofs in mathematics, and it is this type of insight that separates the mathematicians from those that just do math without thinking about what it means. It is a clever use of the fact that adding and subtracting the same value does not change an expression. Also watching Implicit differentiation and logarithmic differentiation has led me to believe this could be proven with natural logarithm but how? That way I can always a problem down into what I know and therefore I can always find the formula for whatever problem I might have. That is, it's easier for me to understand something instead memorizing the entire thing. So replace F(x) and F(x+h) with corresponding terms and you get what Sal got, did I get this right? I really do need to get this right otherwise my brain will refuse to memorize it. Thankfully we already know what the product rule is so we sort of know what we need to add the expression but there has to be a more intuitive way to prove the product rule? Also he's multiplying the entire derivations of 2 functions so why isn't it ((f(x+h)-f(x))*(g(x+h)-g(x)))/h ? Am guessing because the product of 2 derived functions is not the same as derivation of products of functions? This would mean that f(x)*g(x) is now one new function, it's not 2 seperate functions but one as in multiplications where ab is not 2 terms but 1 term, so f(x)*g(x)=F(x), if we evaluate F(x) we get (F(x+h)-F(x))/h and like I said, F(x)=f(x)*g(x). The little trick he uses is not sufficient enough for me to prove to myself the product rule.
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