- Differentiating using multiple rules: strategy
- Differentiating using multiple rules: strategy
- Applying the chain rule and product rule
- Applying the chain rule twice
- Derivative of eᶜᵒˢˣ⋅cos(eˣ)
- Derivative of sin(ln(x²))
- Differentiating using multiple rules
- Product rule to find derivative of product of three functions
How to analyze the structure of an elaborate expression do determine which derivative rules to use, and (not less important) in what order.
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- In the second expression d/dx [sin(x^2 + 5) cos (x)] , wouldn't sin(x^2+5) be a composite function with sin (x) be the outer and x^2 + 5 being the inner functions?(8 votes)
- Yep, Sal even mentions that you would have to solve that part -- if you were actually trying to, for this video only talks about the strategy -- using the chain rule. So you're absolutely right.(16 votes)
- Is the actual solution to the first problem the following?
cos((x^2+5)(cos (x))[2x cos(x)-(x^2+5)sin(x)
I know solutions are not really the point of this video, but I appreciate all the practice I can get(6 votes)
- That's right! I got that answer and thought "No way this is right lol it's way too complex...", so I went to an online calculator and it gave me the same answer! Congrats!(4 votes)
- If a function is the product of two quotients, where would you start? Would it be enough to just do the product rule?(2 votes)
- What you could do is rewrite the function as a single quotient and then use the quotient rule.(3 votes)
- Conceptually i'm comfortable but I am having a computing issue with a practice question. It's not related to multiple rule differentiation, so someone can remove if it shouldn't belong here.
We are doing product rule on three expressions and after differentiating, wind up with this. 2⋅csc(x)⋅sec(x)+2x−csc(x)cot(x)⋅sec(x)+2x⋅csc(x)⋅sec(x)tan(x)
Fine. No problem. But it ends up simplifying to this:
Where did the cot(x) and tan(x) disappear to? I realize this is a trig question and probably a very stupid one, but it's driving me bananas.(1 vote)
- Using the trig definitions, cot(x) = cos(x)/sin(x) and sec(x) = 1/cos(x). Using this we can simplify the third term, -csc(x)cot(x)*sec(x).
We also know that tan(x) = sin(x)/cos(x) and csc(x) = 1/sin(x). We can use this to simplify the final term, 2x*csc(x)*sec(x)*tan(x).
(I apologize for the formatting, but I cannot seem to fix the bold text.)(3 votes)
- d\dx (3x² · √̅5̅x̅+̅3̅ According to Sal, you start from the outside, which would be the product rule. f´(x)g(x) + f(x)g´(x).
= (6x · √̅5̅x̅+̅3̅ )+3x²/2√̅5̅x̅+̅3̅.
Then for the inner part, use the chain rule which is f´(g(x))g´(x).
Will somebody please take a minute and tell me if I got it right, so far? I didn't include the 3x² in the chain rule part. Not too sure about it, and the videos up to now don't elaborate on it.(0 votes)
- That is almost correct. The correct derivative would be 6x√(5x + 3) + 15x²/2√(5x + 3).
The application of the product rule and chain rule were both correct. However, in your final answer, you forgot to multiply by 5, the "g'(x)" in the chain rule. Hope that I helped.(4 votes)
- [Instructor] So I have two different expressions here that I wanna take the derivative of. And what I want you to do is pause the video and think about how you would first approach taking the derivative of this expression and how that might be the same or different as your approach in taking the derivative of this expression. The goal here isn't to compute the derivatives all the way, but really to just think about how we identify what strategies to use. Okay so let's first tackle this one. And the key when looking at a complex expression like either of these is to look at the big picture structure of the expression. So one way to think about it is, let's look at the outside rather than the inside details. So if we look at the outside here, we have the sine of something. So there's a sine of something going on here that I'm going to circle in red or in this pink color. So that's how my brain thinks about it. From the outside I'm like okay, big picture I'm taking the sine of some stuff. I might be taking some stuff to some exponent. In this case, I'm inputting in a trigonometric expression. But if you have a situation like that, it's a good sign that the chain rule is in order. So let me write that down. So we would wanna use in this case the chain rule, C.R. for chain rule. And how would we apply it? Well we would take the derivative of the outside with respect to this inside times the derivative of this inside with respect to x. And I'm gonna write it the way that my brain sometimes thinks about it. So we can write this as the derivative with respect to that something, and I'm just gonna make that pink circle for the something rather than writing it all again, of sine of that something, sine of that something, not even thinking about what that something is just yet, times, times the derivative with respect to x of that something. This is just an application of the chain rule. No matter what was here in this pink colored circle, it might have been something with square roots and logarithms and whatever else, as long as it's being contained within the sine, I would move to this step. The derivative with respect to that something of sine of that something times the derivative with respect to x of the something. Now what would that be tangibly in this case? Well this first part, I will do it in orange, this first part would just be cosine of x squared plus five times cosine of x. So that's that circle right over there. Let me close the cosine right over there. And then times the derivative with respect to x, times the derivative with respect to x, of all of this again, of x squared plus five times cosine of x. And then I would close my brackets. And of course I wouldn't be done yet, I have more derivative taking to do. Here now I would look at the big structure of what's going on, and I have two expressions being multiplied. I don't have just one big expression that's being an input into like a sine function or cosine function or one big expression that's taken to some exponent. I have two expressions being multiplied. I have this being multiplied by this. And so if I'm just multiplying two expressions, that's a pretty good clue that to compute this part, I would then use the product rule. And I could keep doing that and compute it, and I encourage you to do so, but this is more about the strategies and how do you recognize them. But now let's go to the other example. Well this looks a lot more like this step of the first problem than the beginning of the original problem. Here I don't have a sine of a bunch of stuff or a bunch of stuff being raised to one exponent. Here I have the product of two expressions just like we saw over here. We have this expression being multiplied by this expression. So my brain just says okay I have two expressions, then I'm going to use the product rule. Two expressions being multiplied, I'm going to use the product rule. If it was one expression being divided by another expression, then I would use the quotient rule. But in this case it's going to be the product rule. And so that tells me that this is going to be the derivative with respect to x of the first expression, just gonna do that with the orange circle, times the second expression, I'm gonna do that with the blue circle, plus the first expression, not taking its derivative, the first expression, times the derivative with respect to x of, of the second expression. Once again here, this is just the product rule. You can substitute sine of x squared plus five where you see this orange circle, and you can substitute cosine of x where you see this blue circle. But the whole point here isn't to actually solve this or compute this, but really to just show how you identify the structures in these expressions to think about well do I use the chain rule first and then use the product rule here? Or in this case do I use the product rule first? And even once you do this, you're not going to be done. Then to compute this derivative, you're going to have to use the chain rule, and you'll keep going until you don't have any more derivatives to take.