![]() 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). This expands to 3072x^5 - 2880x^4 + 864x^3 - 81x^2.The little trick he uses is not sufficient enough for me to prove to myself the product rule. We have already found f'(g(x)) and g'(x) separately now we just have to multiply them to find the derivative of the composite function. Since g(x) = 8x^2-3x, we know by the power rule that g'(x) = 16x-3.Īccording to the chain rule, as we saw above, the derivative of f(g(x)) = f'(g(x)) g'(x). The next step is to find g'(x), the derivative of g. The derivative of f(x) is 3x^2, which we know because of the power rule. The first step is to take the derivative of the outside function evaluated at the inside function. We can apply the chain rule to your problem. In plain (well, plainer) English, the derivative of a composite function is the derivative of the outside function (here that's f(x)) evaluated at the inside function (which is (g(x)) times the derivative of the inside function. ![]() To differentiate a composite function, you use the chain rule, which says that the derivative of f(g(x)) = f'(g(x)) g'(x). ![]() ![]() That's the function you have to differentiate. Let's call the two parts of the function f(x) and g(x). It's not as complicated as it looks at a glance! The trick is to use the chain rule. ![]()
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