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AP®︎/College Calculus AB
Unit 3: Lesson 7
Selecting procedures for calculating derivatives: multiple rules- 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
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Applying the chain rule twice
AP.CALC:
FUN‑3 (EU)
Worked example applying the chain rule twice.
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I'm having trouble with applying the 2nd chain rule. I have no idea which side I'm supposed to use the chain rule. For example, in the quotient rule, d/dx [u(v(x))/w(x)] it will become d/dx [u( v(x) )] * w(x) - u( v(x) ) * w'(x)/ ( w(x) )^2. Why is the chain rule used for the first part of the difference? Why can't it be used on the 2nd part of the difference?(3 votes)
I am struggling with this also. The video is about applying the chain rule twice, there may be other ways to get the answer, but first I want to understand how to apply the chain rule twice, which can be confusing.
Where Sal draws a parenthesis and says something goes in there seems to gloss over a little bit how the chain rule entity is identified and then goes in there, and next he says to apply the product rule. So, I think what he is saying is let u = everything which is inside the parentheses, then we have y = u^3, dy/du = 3 u^2 = 3 times (everything inside the parentheses) squared. Applying the product rule is the easy part.
He then goes on to apply the chain rule a second time to what is inside the parentheses of the original expression. And finally multiplies the result of the first chain rule application to the result of the second chain rule application.
Earlier in the class, wasn't there the distinction between outside and inside of an expression? Here we apply the chain rule to the outside first (the cube function), and secondly we apply it to the inside function, the sin(x^2).
I've spent a lot of time on this and now I think it is starting to make sense.(3 votes)
Can someone help me. It is not in the video but a question after. y = (sqrt2x^2)/Cos(x) , find dy/dx . My first question is why can we not convert sqrt2x^2 to (2x^2)^½ like with indices rules where square root is same as to the power of ½ .. ? I know I am missing something but I just cannot see what it is. Getting past that and going on in the solution using the chain rule when everything is in place -(Sqrt2x^2)(-sin x) becomes +4x^2(six) where is the 4x^2 coming from ?(3 votes)
@2:24shouldnt it be 3 sin^2(x^4)(2 votes)
No, this is because, 3(sin(x^2))^2=3(sin^2(x^4)) not 3(sin^2((x^2)^2)).(1 vote)
At2:35why doesn't he also take the derivative of 2x and multiply the whole thing by 2? Shouldn't he just keep taking the derivative of what results from the previous derivation until the derivative is zero?(2 votes)
No. He was using the chain rule to evaluate the compositions of functions. However, x^2 can be evaluated using the power rule, so he did not need to differentiate the result.(1 vote)
what's the difference between d/dx and dy/dx? why does it sometimes mean derivative and is sometimes used as a constant?(1 vote)
d/dx is the operator meaning "take the derivative of, with respect to x"
dy/dx is a term indicating that the derivative of y was taken with respect to x(1 vote)
- I am really confused. At2:18, 3(sin(x^2))^2 is directly converted into 3sin^2(x^2), I know it's correct. But isn't 3(sin(x^2))^2 a composite function? The inner function g(x)= sin(x^2), the outer function f(x)= 3(x^2).
How can we just multiply the square into the function in the bracket?(1 vote)
Yes, it is a composite function. But I don't get what you meant by 'multiply the square into the function in the bracket.' You're just putting the sin(x^2) instead of x of f(x).(1 vote)
At1:07why does Sal switch from taking the derivative of y with respect to x to taking the derivative of x with respect to x?(1 vote)
At1:07he uses the chain rule to differentiate a "new" function which isn't y. This means he has to write d/dx instead of dy/dx. dy/dx just means "the derivative of y with respect to x". Since y is out of the equation here, we just use d/dx.(1 vote)
2:24Video transcript
- [Instructor] Let's say that Y is equal to sin of X
squared to the third power, which of course we could also write as sin of X squared to the third power and what we're curious about is what is the derivative
of this with respect to X? What is DY/DX which we
could also write as Y prime? Well, there's a couple of
ways to think about it. This isn't a straightforward
expression here but you might notice that I have something being raised to the third power, in fact, if we look at the
outside of this expression we have some business in here that's being raised to the third power. And so, one way to tackle this is to apply the chain rule. So, if we apply the chain rule it's gonna be the
derivative of the outside with respect to the inside or the something to the third power, the derivative of the
something to the third power with respect to that something. So, it's going to be three
times that something squared times the derivative with respect to X of that something, in this case, the something is sin, let me write that in the blue color, it is sin of X squared. It is sin of X squared. No matter what was inside
of these orange parentheses I would put it inside of
the orange parentheses and these orange brackets right over here. We learned that in the chain rule. So, let's see, we know
this is just a matter of the first part of the expression is just a matter of
algebraic simplification but the second part we need
to now take the derivative of sin of X squared. Well, now we would want to
use the chain rule again. So, I'm going to take the derivative, it's sin of something, so this is going to be,
the derivative of this is gonna be the sin of something with respect to something, so that is cosine of that something times the derivative with respect to X of the something. In this case, the
something is our X squared and of course, we have
all of this out front which is the three times sin of X squared, I could write
it like this, squared. Alright, so we're getting close. Now we just have to
figure out the derivative with respect to X of X squared and we've seen that many times before. That, we just use the power rule, that's going to be two X. Two X and so, if we
wanted to write the DY/DX, let me get a little bit
of a mini drum roll here, this shouldn't take us too long, DY/DX, I'll multiply the
three times the two X which is going to be six X, so I've covered those so far times sin squared of X squared, times sin squared of X squared, times cosine of X squared. And we are done applying the
chain rule multiple times.